Let $(M,d_M)$ be a separable metric space. That is, there is a countable subset $S\subset M$ whose closure in $M$ is $M$ itself, $\text{cl}_M(S)=M$.
Now suppose there is a homeomorphism $f:M\to N$. Must $N$ be separable?
My thoughts are as follows. Since $S$ is countable, so is $F=f(S)$. Using this, I broke up the problem into four cases, the last two of which I have been unable to complete.
Case 1:
If $S$ is closed in $M$, then $\text{cl}_M(S)=S=M$ and $F=f(S)=f(M)=N$, meaning that $\text{cl}_{N}(F)=N$. Thus $N$ is separable.
Case 2:
If $S$ is clopen (closed and open) in $M$, it is closed in $M$, and $N$ is separable.
Case 3:
If $S$ is open in $M$, $F$ is open in $N$, but this doesn't seem to help.
Case 4:
If $S$ is neither open nor closed in $M$, I have no idea what to do.
There must be a better way to go about this. Could I have some help? Thanks.
A homeomorphism is a bijection that takes open sets to open sets, closed sets to closed sets, so it follows that it also respects closure, therefore $$\mathrm{cl}_N(f(S)) \ =\ f(\mathrm{cl}_M(S)) \ =\ f(M) \ =\ N$$