Understanding Continuity defined by closure Proposition:

Let $f; T_1 \to T_2$ be a continuous map between topological spaces.
Then for every $H$, $f(\overline{H}) \subset \overline{f(H)}$.

Proof 1 (from Continuity defined by closure):

Let $y \in f(\overline{H}).$ There exists $x \in \overline{H},$ with
$y=f(x)$. Let $U$ be any open set of $T_2$ such that $y \in U$. By
continuity, $f^{-1}U$ is open in $T_1$, and $x$ belongs to $f^{-1}U$.
Thus $(f^{-1}U) \cap H$ is not empty, since $x$ is a closure point for
$H$.
Hence $U \cap f(H) \supset f((f^{-1}U) \cap H) \not= \emptyset,$ and
$y \in \overline{f(H)}.$

I need help understanding this Necessary Condition. It is one direction of Proof 1 of Continuity defined by closure.
Why $f^{-1} [ U] \cap H \ne \emptyset$ as $x \in \text{cl}(H)$?
Why $\emptyset \neq f[f^{-1}[U]\cap H] \subseteq U \cap f[H]$?
And why $y \in \text{cl}(f[H])$?
Any help appreciated.
 A: For why $f^{-1}[U] \cap H \neq \emptyset$ as $x \in \text{cl}(H)$: One has a theorem that states $x \in \text{cl}(H)$ if and only if for every open set $U$ containing $x$, we have $U \cap H \neq \emptyset$. This is not too difficult to prove by using the contrapositive statement: $x \notin \text{cl}(H)$ if and only if there is some open set $U$ containing $x$ such that $U \cap H = \emptyset$. Suppose $x \notin \text{cl}(H)$, then $T_1 \backslash \text{cl}(H)$ is an open set containing $x$ that does not intersect $H$. On the other hand, if there exists an open set $U$ containing $x$ that does not intersect $H$, then $T_1 \backslash U$ is a closed set containing $H$, and by definition of closure, $\text{cl}(H)$ is a subset of every closed set that contains $H$, so $x \notin \text{cl}(H)$. So the statement that $f^{-1}[U] \cap H \neq \emptyset$ as $x \in \text{cl}(H)$ is true because $f^{-1}[U]$ is an open set containing $x$.
For why $\emptyset \neq f[f^{-1}[U] \cap H] \subseteq U \cap f[H]$: This follows because for any map of sets $f:X \rightarrow Y$, and any two subsets $A,B \subseteq X$, we have $f(A \cap B) \subseteq f(A) \cap f(B)$. Let $y \in f(A \cap B)$, then there is some $x \in A \cap B$ such that $f(x) = y$. Since $x \in A$, we have $y = f(x) \in f(A)$, and since $x \in B$, we also have $y = f(x) \in f(B)$. Hence $f(A \cap B) \subseteq f(A) \cap f(B)$.
For why $y \in \text{cl}(f[H])$: This again follows from the statement in the first paragraph. $U$ is an arbitrary open set containing $y$, and we have $U \cap f[H] \neq \emptyset$, and so it follows that $y \in \text{cl}(f[H])$.
A: Instead of a “pointwise” proof, I would use the “set-wise” proof for that proposition:
$\overline{f[H]}$ is closed in $Y$ so by continuity of $f$
$$f^{-1}[\overline{f[H]}] \text{ is closed in } X\tag{1}$$
And as $f[H] \subseteq \overline{f[H]}$ trivially by simple set theory we have
$$H \subseteq f^{-1}[f[H]] \subseteq f^{-1}[\overline{f[H]}]\tag{2}$$
And as the right hand set of $(2)$ is closed and $\overline{H}$ is the smallest closed superset of $H$ we have
$$\overline{H} \subseteq f^{-1}[\overline{f[H]}]\tag{3}$$
And from $(3)$ by simple set theory it follows that
$$f[\overline{H}] \subseteq \overline{f[H]}$$ and we’re done.
A: The use of elements is unnecessary. You certainy know that a map is continuous if and only if $f^{-1}(C) \subset T_1$ is closed for all closed $C \subset T_2$.
Let $H \subset T_1$. Then $H \subset f^{-1}(f(H) \subset f^{-1}(\overline{f(H)})$. Since $f^{-1}(\overline{f(H)})$ is closed, we see that $\overline{H} \subset f^{-1}(\overline{f(H)})$ which implies $f(\overline{H}) \subset f(f^{-1}(\overline{f(H)})) \subset \overline{f(H)}$.
