Let X and Y Hausdorff normal topological spaces, and let N,M dense subspaces of X,Y, respectivaly.
Let f a homeomorphism between N and M. Is true that exists an continuos extension F (between X and Y) of f?
F may also be a homeomorphism between X and Y?
The answer is No, even if $X$ and $Y$ are compact.
Let $X = S^1$, $M$ be $S^1\setminus\{p\}$, that is , $M$ is $S^1$ with a point removed. Let $Y =[0,1]$ and $N = (0,1)$.
Obviously, $M$ and $N$ are homeomorphic, so let $f:M\rightarrow N$ be such a homeomorphism.
Then there is no continuous extension $F:X\rightarrow Y$. This follows because $F(p)$ is forced to simultaneously be $0$ and $1$ by continuity.
This example can also be used to show that $F$, if it exists, needs not be a homeomorphism: Let $g:N\rightarrow M$ be a homeomorphism. Then there is a (unique) extension $G$ and $G$ has the property that $G(0) =G(1) = p$. In particular, $G$ is not injective, so not a homeomorphism.
(On the other hand $G$ will be surjective, and this must happen in the compact case: $G(Y)$ must be compact, being the image of a compact space, so is closed because $X$ is Hausdorff. But $G(Y)$ contains the dense set $M$, so $X = \overline{M}\subseteq \overline{G(Y)} = G(Y)$, so $X\subseteq G(Y)$.)