Maybe not as hard as you see

[Asif Hossain]

Prove that any closed loop is inscribed in a square.(It is not the unsolved topology problem rather a brain teaser)
Disclaimer: I don't know how to prove it formally but the intuitive proof(key) is enough.
Hint:

It is easy to prove that every closed loop is inscribed in a rectangle then try to mess with the rectangle a little bit and hopefully you should be done.

[Asif Hossain]

Prove that any closed loop is inscribed in a square.(It is not the unsolved topology problem rather a brain teaser)
Disclaimer: I don't know how to prove it formally but the intuitive proof(key) is enough.
Hint:

It is easy to prove that every closed loop is inscribed in a rectangle then try to mess with the rectangle a little bit and hopefully you should be done.

Still no answer

[Asif Hossain]

BUMPTY BUMPTY BUMP Why no body trying this beautiful problem?

[Anindya Biswas]

Let's imagine a big rectangle $ABCD$ (named in clockwise direction) inside of which a closed loop is lying.
Now start moving the side $AB$ towards $CD$ until it touches the closed loop. Do this for every sides, pull them towards the opposite side until they touches the loop.
Following this algorithm, we will find a rectangle in which the loop is inscribed.
Now if it is a square, we are done. If $AB>BC$, start rotating the rectangle counterclockwise while keeping it's sides tangent to that loop (this may change it's side lengths.)
After rotating $90$ degrees, It becomes the same rectangle we started with, but now point $A$ goes to the place where point $B$ initially was, and other point changed their positions accordingly.
Therefore, now we have $BC>CD\Rightarrow BC>AB$
So, we started with $AB>BC$ and ended up with $AB<BC$ after a continuous change.
So, somewhere in between we must get a rectangle where $AB=BC$. And that will be a square.

[Anindya Biswas]

Does this matches your reasoning? I highly doubt if "rotating the rectangle and keeping it's side tangent to the loop" is a continuous action or not.

[Asif Hossain]

Let's imagine a big rectangle $ABCD$ (named in clockwise direction) inside of which a closed loop is lying.
Now start moving the side $AB$ towards $CD$ until it touches the closed loop. Do this for every sides, pull them towards the opposite side until they touches the loop.
Following this algorithm, we will find a rectangle in which the loop is inscribed.
Now if it is a square, we are done. If $AB>BC$, start rotating the rectangle counterclockwise while keeping it's sides tangent to that loop (this may change it's side lengths.)
After rotating $90$ degrees, It becomes the same rectangle we started with, but now point $A$ goes to the place where point $B$ initially was, and other point changed their positions accordingly.
Therefore, now we have $BC>CD\Rightarrow BC>AB$
So, we started with $AB>BC$ and ended up with $AB<BC$ after a continuous change.
So, somewhere in between we must get a rectangle where $AB=BC$. And that will be a square.

After rotating 90 degrees or 180 degrees? (If we label one side as blue and one side as brown after rotating 180 degrees the |brown side| becomes |blue| and |blue side| becomes |brown| )

[Anindya Biswas]

Let's imagine a big rectangle $ABCD$ (named in clockwise direction) inside of which a closed loop is lying.
Now start moving the side $AB$ towards $CD$ until it touches the closed loop. Do this for every sides, pull them towards the opposite side until they touches the loop.
Following this algorithm, we will find a rectangle in which the loop is inscribed.
Now if it is a square, we are done. If $AB>BC$, start rotating the rectangle counterclockwise while keeping it's sides tangent to that loop (this may change it's side lengths.)
After rotating $90$ degrees, It becomes the same rectangle we started with, but now point $A$ goes to the place where point $B$ initially was, and other point changed their positions accordingly.
Therefore, now we have $BC>CD\Rightarrow BC>AB$
So, we started with $AB>BC$ and ended up with $AB<BC$ after a continuous change.
So, somewhere in between we must get a rectangle where $AB=BC$. And that will be a square.

After rotating 90 degrees or 180 degrees? (If we label one side as blue and one side as brown after rotating 180 degrees the |brown side| becomes |blue| and |blue side| becomes |brown| )

After rotating $180$ degrees, the side lengths stays the same with what they started with. This shouldn't guarantee that at some angle the smaller side and the bigger side actually becomes same. In the other hand, rotating $90$ degrees leaves the smaller side bigger and the bigger side smaller which guarantees that at some point they were same.

[Anindya Biswas]

Does this matches your reasoning? I highly doubt if "rotating the rectangle and keeping it's side tangent to the loop" is a continuous action or not.

.
The word "tangent" here is somewhat inappropriate since the curves is not necessarily smooth. In that case, I am referring to a line that contains at least one point of the loop, and every point on the loop is either on that line, or a specific side of that line. No point of that loop can cross the line.

[emeryhen121]

There are two possibilities
Possibility A: The four vertices of a square must be present in any closed and limited set that divides the plane into more than one component.
Possibility B: Inscribed squares are not present in the majority of simple closed curves.
Stormquist’s Theorem states “If the simple closed curve J is "nice enough" then it has an inscribed square”. "Nice enough" indicates that there must be a coordinate system for the plane in which some section of the curve containing P is the graph y = f(x) of a continuous function for each point P on the curve. Stromquist's theorem is good enough to cover any curve you could draw with a pencil and paper. In other words, this theorem essentially answers the informal version of our initial question: any curve drawn on a piece of paper will have an inscribed square.
According to multiple websites available on Google, the following theorems could be concludes
Theorem 1: Every simple closed curve that is symmetric about the origin has an inscribed square.
Theorem 2: Every simple closed curve has lots of inscribed parallelograms and lots of inscribed rhombuses.
Theorem 3: Every simple closed curve has at least one inscribed rectangle
Theorem 4: Extending Theorem D, J has so many inscribed triangles similar to T that the vertices of all these inscribed triangles are "dense" in the curve J.
These all theorems are available on multiple research papers available easily online. You can just write these keywords and find them to get a deep knowledge of the topic.

Hope it helps!