Question about a proof of $f(\overline A)\subseteq\overline{f(A)}$ for all $A\subset{\rm dom}(f)$ implies that $f$ is continuous 
If $f(\overline A)\subseteq\overline{f(A)}$ for all $A\subset{\rm dom}(f)$ then $f$ is continuous.

I know there are others questions about this topic but I want to know what more I need to conclude my proof, if possible. The way Im trying to prove this is different to what I see in other answers about this topic. The context of this proof is
$$f:X\to Y$$
where $X$ and $Y$ are metric spaces.
Then to proof that if $f(\overline A)\subseteq\overline{f(A)}$ then $f$ is continuous I will reason with convergent sequences.
We have that for any sequence $(x_n)\to x$ with $x_n\in A$ for all $n\in\Bbb N$ then $x\in\overline A$.
Then observe that the statement $f(\overline A)\subseteq\overline{f(A)}$ implies that if $f(x_n)\in f(A)$ then $f(x)\in\overline{f(A)}$, i.e. the image of the limit of any convergent sequence in $A$ is defined in the closure of the image of $A$, in other words
$$f(\lim x_n)=f(x)\in\overline{f(A)}$$
But I dont know if I can conclude that
$$\lim f(x_n)=f(x)$$
what is the definition of continuity on metric spaces. In other words: what I need to conclude
$$\lim f(x_n)=f(\lim x_n)$$
in this context?
 A: The proof in topological spaces:
Assume that $f$ is not continuous, that is there exists some open set $V\subset Y$ with $U:=f^{-1}(V)$ is not open. Let $A = X\setminus U$. 
Notice that for every $x\in X$ we have
$$ x \in A = X\setminus f^{-1}(V) \iff x\notin f^{-1}(V) \iff f(x) \notin V,$$
that is $f(A) \subseteq Y\setminus V$. 
Now, as $U$ is not open, there exists a $x\in U \cap \bar A$ and we have


*

*$x\in U = f^{-1}(V) \iff f(x) \in V$ and

*$x\in \bar A \implies f(x) \in f(\bar A) \subseteq \overline{f(A)} \subseteq Y \setminus V$,


which is a contradiction.
Old answer:
Proof only for metric spaces:
Assume the converse: Let $x_n\to x$ and $f(x_n)\not\to f(x)$. By going to a subsequence we may assume that there exists a $\epsilon > 0$ such that $|f(x_n) - f(x)| \ge \epsilon$ holds for all $n$. 
On other hand we have
$$ f(x) \in \overline{\{ f(x_n) \mid n\in \mathbb N \}}. $$
That is, there exists a sequence $f(x_{n_k}) \to f(x)$, which is a contradiction.
A: This is too long for a comment, but hopefully a useful addition.
The topological proof given by user251257 is neat, but perhaps worth some expansion. 
If $U$ is not open then $X \setminus U$ is not closed and therefore doesn't contain all its limit points. So there is a limit point $x$ of $X \setminus U$ in $X \setminus (X \setminus U) = U$.
$x $ and other limit points of $X \setminus U$ are in the closure of $X\setminus U $, so as stated  there is $x \in U \cap \bar A$ and it follows (1) $f(x) \in V$. 
For (2), $ \overline{f(A)} $, the closure of $f(A)$ is a subset of every closed set that contains $f(A)$; since $V$ is open then $Y \setminus V$ is closed, and as shown initially $f(A) \subset Y \setminus V$: therefore $Y \setminus V $ is a closed set containing $f(A)$ and therefore contains $ \overline{f(A)} $. 
(Since it took me a while to work this out I thought I'd share it).
