For every subset $A$ of $X$, $f(\overline{A})$ is a subset of $\overline{f(A)}$

Question

looking at the image above, the line $3 \to 4$, I can't seem to understand why $f(x)$ belongs to the closure of $f(A)$ and how it implies that $f(\overline{A})$ is a subset of $\overline{f(A)}$.

Answer 1

1. The author took a $x\in\overline A$ and proved that, for any neighborhood $V$ of $f(x)$, $V\cap f(A)\neq\emptyset$. But this means that $f(x)\in\overline{f(A)}$.
2. The author proves that, for each $x\in\overline A$, $f(x)\in\overline{f(A)}$. But this means that $f\left(\overline A\right)\subset\overline{f(A)}$.

Answer 2

You started with an arbitary point of $\overline{A}$. You then show

Every neighbourhood of $f(x)$ intersects $f[A]$.

This is the definition of $f(x) \in \overline{f[A]}$. And so $f[\overline{A}] \subseteq \overline{f[A]}$, as a point $f(x)$ as you started with is really an arbitrary point of $f[\overline{A}]$.