I want to show:
Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Then is $\overline{Y}$ closed.
For my proof I want to write $X\setminus\overline{Y}$ as union of open sets.
Sounds like a good approach!
Proof:
It is $\overline{Y}=Y\cup\partial Y$.
Consider $X\setminus(Y\cup\partial Y)=\overline{Y}^c$.
Here, I might instead say something like this: "By definition, $\overline Y=Y\cup\partial Y.$ We show that $X\setminus\overline Y$ is a union of open sets, so is open."
Let $x\in X\setminus(Y\cup\partial Y)$. Hence $x\notin Y$ and $x\notin\partial Y$.
Therefor for every neighborhood $U_x$ of $x$ it is $U_x\cap Y=\emptyset$.
This isn't quite right. Rather, by definition of $\partial Y,$ since $x\notin Y,$ then there exists a neighborhood $U_x$ of $x$ such that $U_x\cap Y=\emptyset.$
It's also a bit awkward, and there's a misspelling. I might say something like this: "Let $x\in X\setminus\overline Y.$ Since $\overline Y=Y\cup\partial Y,$ then by DeMorgan's Laws, $X\setminus\overline Y=(X\setminus Y)\cap(X\setminus\partial Y),$ so $x\notin Y$ and $x\notin\partial Y.$ Therefore, by definition of $\partial Y,$ there exists some neighborhood $U_x$ of $x$ such that $U_x\cap Y=\emptyset.$ Since $x\in X\setminus\overline Y$ was arbitrary, then such a $U_x$ exists for each such $x.$"
Hence $U_x\subseteq X\setminus Y$ and $\overline{Y}^c\subseteq\underbrace{\bigcup_{x\in\overline{Y}^c} U_x}_{\text{open}}\subseteq X\setminus Y$.
Nicely done! There's really no purpose to mentioning that the union is open right now, though. I'd wait until the end. (See what I do there.)
Now I want to show, that $\overline{Y}^c\supseteq\bigcup_{x\in\overline{Y}^c} U_x$.
Let $x'\in\bigcup_{x\in\overline{Y}^c} U_x$.
I have to show, that $x'\notin Y$ and $x'\notin\partial Y$.
You can certainly proceed in this way, though since $X\setminus\overline Y=(X\setminus Y)\cap(X\setminus\partial Y),$ and since you've already shown that $$X\setminus\overline{Y}\subseteq\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus Y,$$ then you need only show that $$\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus\partial Y,$$ meaning that you only have to show $x'\notin\partial Y.$
Since $x'\notin\bigcup_{x\in\overline{Y}^c} U_x\subseteq X\setminus Y$ is $x\notin Y$.
This doesn't make sense. It seems like you're trying to say that, since $x'\in\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus Y,$ then $x'\notin Y.$ However, as I said, we don't even need to say this.
Suppose $x'\in\partial Y$.
That's what I'd do!
Then holds for every neighborhood $U_{x'}$ of $x'$, that $Y\cap U_{x'}\neq\emptyset$ and $U_{x'}\cap (X\setminus Y)\neq\emptyset$.
Which contradicts, that $U_{x'}\subseteq X\setminus Y$.
You've got the right idea, but it seems that you're trying to let $U_{x'}$ be simultaneously arbitrary and a specific counterexample. Instead, I'd say something like this: "By definition of $\partial Y,$ this means that for every neighborhood $U$ of $x',$ we have $U\cap Y\neq\emptyset.$ However, since $x'\in\bigcup_{x\in X\setminus\overline Y}U_x,$ then we have that $U_{x'}$ is a neighborhood of $x'$ disjoint from $Y,$ yielding the desired contradiction."
We condlude, that $\overline{Y}^c=\bigcup_{x\in\overline{Y}^c} U_x$ open.
Hence $\overline{Y}$ is closed.
Here, I'd just say (if you lead off as I did by announcing your intention) something like: "We conclude that $$X\setminus\overline Y\subseteq\bigcup_{x\in X\setminus\overline Y}U_x\subseteq(X\setminus Y)\cap(X\setminus\partial Y)=X\setminus\overline Y,$$ so that $X\setminus\overline Y=\bigcup_{x\in X\setminus\overline Y}U_x.$ As a union of the open sets $U_x,$ we have that $X\setminus\overline Y$ is open, as we set out to show."
Let me know if you have any questions about my answer, or if you just want to bounce your phrasing adjustments off somebody.
Best Answer
$f[B]=f[f^{-1}[A]] \subseteq A$, not $=A$ (unless $f$ is surjective), but that's a minor thing; you only need the inclusion anyway.
The first part can be done easier: As $\overline{f[A]}$ is closed, $f^{-1}[\overline{f[A]}]$ is closed by continuity and contains $A$, i.e. $$A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$$
so $$f[\overline{A}] \subseteq \overline{f[A]}$$ is immediate. This does require you know that $f$ continuous means that the inverse images of closed sets are closed. (As for open sets.)