Given a continuous and closed map $f$ between two metric space $E, F$, prove that $f(\overline{A}) = \overline{f(A)}$ for all $A \subset E$ I already know of this question ($f$ is continuous and closed $\Longleftrightarrow \overline{f(E)} = f(\overline{E})$ for all $E \subseteq M$) which is quite the same as mine.
But I don't understand some chunks of the proof.
First, let $x \in \overline{A}$, we can build a sequence $u_n \in A^{\mathbb{N}}$ which converges to $x \in \overline{A}$.
Also, we already know by definition that $f(x) \in f(\overline{A})$.
Then, as $f$ is continuous, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(x)$, so we deduce $f(x) \in \overline{f(A)}$ (I don't fully understand why we can deduce this.)
Now, we have $f(\overline{A}) \subset \overline{f(A)}$.
Finally, as $f(\overline{A})$ is a closed set because $f$ is a closed map and $\overline{A}$ is a closed set by definition.
We can deduce that $f(A) \subset f(\overline{A})$ and as $f(\overline{A})$ is closed, $f(A)$ being included in it indicate that we also have smallest closed set containing $f(A)$ in $f(\overline{A})$, at least, I understand this. As a result, we can deduce that $\overline{f(A)} \subset f(\overline{A})$, but why?
I am a bit lost and new to closed / open sets, sorry for the formatting.
 A: 
Then, as $f$ is continuous, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(x)$, so we deduce $f(x) \in \overline{f(A)}$ (I don't fully understand why we can deduce this.)

Let $y = f(x)$ and $v_n = f(u_n)$ for each $n$, then what is stated there is that the sequence $v_n$ converges to $y$ (by definition of continuity).
On the other hand, each $v_n$ is in $f(A)$ because each $u_n$ is in $A$, and the limit of any sequence in $f(A)$ lies in $\overline{f(A)}$ (can you see why is this the case?), therefore $y$ itself, being the limit of elements in $f(A)$, belongs to $\overline{f(A)}$. In other words, $f(x)\in \overline{f(A)}$.

We can deduce that $f(A) \subset f(\overline{A})$ and as $f(\overline{A})$ is closed, $f(A)$ being included in it indicate that we also have smallest closed set containing $f(A)$ in $f(\overline{A})$, at least, I understand this. As a result, we can deduce that $\overline{f(A)} \subset f(\overline{A})$, but why?

Did you understand this claim: we also have smallest closed set containing $f(A)$ in $f(\overline{A})$? If you do, then you're done because by definition (or, as a property depending on your definition of closure) the set $\overline{f(A)}$ is precisely "smallest closed set containing $f(A)$".
A: The topological def'n is that $f:X\to Y$ is continuous iff $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y.$ Regardless of whether the topologies can be defined in terms of sequences. There are many consequences of the continuity of $f$ that also imply the continuity of $f,$ and could be taken as alternate, equivalent def'ns. One of these is :  $$f(Cl_X(A))\subset Cl_Y(f(A)) \;\text {for all } A\subset X.$$
(i). Suppose $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y$.  Then for $A\subset X,$ the set $C=Y \backslash Cl_Y(f(A))$ is open in $Y,$ so $f^{-1}C$ is open in $X,$ and $(f^{-1}C)\cap A=\phi.$ 
Now for any open set $P,$ if $P\cap Q=\phi$ then $P\cap \bar Q=\phi.$ So $(f^{-1}C)\cap \bar A=\phi.$ Therefore  $f(\bar A)\subset Y \backslash C=Cl_Y(f(A)).$
(ii). I will leave the converse of (i) to you.
By (i), if $f:E\to F$ is a continuous closed map and $A\subset E$ then    $$f(\bar A)\subset \overline {f(A)}\subset \overline {f(\bar A)} \;\text { by continuity of } f$$ $$\text { and }\quad  f(\bar A)= \overline {f(\bar A)}\; \text { by closedness of  } f$$ $$\text {so }\quad \overline {f(A)}=f(\bar A).$$
A: $f: X \to Y$ is continuous iff $$\forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$$
See this question and its answers.
$f: X \to Y$ is closed iff $$\forall A \subseteq X: \overline{f[A]} \subseteq f[\overline{A}]$$
If $f$ is closed then pick any $A \subseteq X$ then $f[A] \subseteq f[\overline{A}]$ as $A \subseteq \overline{A}$. But $f[\overline{A}]$ is closed as the image of a closed set under $f$ so $\overline{f[A]} \subseteq \overline{f[\overline{A}]} = f[\overline{A}]$.
Now suppose the inclusion property holds and let $A$ be closed in $X$. Then $A = \overline{A}$ and we get $(f[A] \subseteq) \overline{f[A]} \subseteq f[A]$, so $f[A]$ equals its closure and so $f[A]$ is closed.
From these two equivalences we get the characterisation of closed and continuous functions.
Nothing we do here uses metrics, just general properties of closure and continuity.
A: $\overline {A} $ is closed, not because $A $ is open but if $(a_n) $ is a sequence of elements from $A $, which converges to $a $, then
$$\forall \epsilon>0, \exists N\in \Bbb N  \;\;:\;\;
d (a,a_N)<\epsilon $$
and this means that $a\in \overline {A} $.
For example
$A=(0,1] $ is not open but
$\overline {A}=[0,1] $ is closed.
