Let M and N be surfaces in $R^3$ such that M is contained in N.
If M is compact and N is connected, prove that M=N.
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I thought intuitively the compactness means that M must be "rolled" until it meets itself, like sphere or torus.
This means that the 3rd dimension is involved and M is closed. But M is contained in a surface N. Also, N is connected so that it is composed of only one chunk.
Thus N must be M.
However I can't verify whether this idea is correct or wrong.
Furthermore I can't prove this in rigorous way.
It seems that I have to use neighborhoods to prove this problem...
I need help ;(
The answer is essentially already given in the comments. Let me elaborate a little bit.
Your intuition is basically correct. Because you have a closed surface, there is no "boundary", which means that there is no way to extend your surface (without making it disconnected). But how to make this precise?
As is often the case when it comes to the notion of connectedness, you will want to prove that $M$ is both open and closed in $N$. Being closed comes for free since $M$ is compact.
To show that $M$ is open, the idea is also simple. Every point in $M$ has a neighbourhood that "looks like" a disc in $\mathbb{R}^2$. Since $N$ also looks locally like $\mathbb{R}^2$, this disc "should" be open in $N$, so that $M$ (as a union of open sets) is open in $N$, and we are done.
To make this precise, we should first clarify your assumptions: What does "looks like" mean here? If your surfaces are assumed to be smooth, then "looks like" means "is diffeomorphic to", and you can use the inverse function theorem, as suggested by JonSK.
If your surface is only a topological surface, then you cannot use the inverse function theorem, because "looks like" then means "is homeomorphic to". But you can still use the invariance of domain theorem, as stated by Anubhav.K.
Note that none of this really requires that your surfaces are embedded in $\mathbb{R}^3$, or that they are two-dimensional. So you see that, more generally: If $M\subset N$, where $M$ is compact, $N$ is connected, and $M$ and $N$ are both $k$-dimensional manifolds, then $M=N$.