Proof that the boundary of the square is not an embedded submanifold There are other questions about this on MSE, but none I've found gives a direct proof of the fact that the (boundary of the) square of, say, length $2$ centered at the origin is not an embedded submanifold of $\mathbb{R}^2$.
I do not see how to give a formal proof that the square cannot be the image of a smooth embedding into $\mathbb{R}^2$. The idea is to show that smoothness would break down at the corners, but I would appreciate seeing a somewhat fleshed out argument.

I will try to give the steps.
First, show that if it is a smooth manifold embedded in $\mathbf{R}^2$, it has to be of dimension $1$.
Next show that any for $1$-dimensional smooth manifold embedded in $\mathbf{R}^2$, locally either the $x$-coordinate or the $y$-coordinate will give the coordinate charts. 
Then see that that's not possible for the corner points, cause any open neighborhood around a corner point will have points with same $x$-coordinates, same argument shows that $y$-coordinates also can't give a coordinate chart.
