Proof that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$ for open $H$ This appears in W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st ed. 1975), chapter 3, exercise 3.9: 27 (iii).
"If $H$ is open in $T$ and $K \subseteq T$ then:
$$H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$$"
(where $\operatorname{cl}$ denotes closure and $T$ is a topological space.)
It's obvious that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$ but then it does not follow that $\operatorname{cl}(H) \cap \operatorname{cl}(K) = \operatorname{cl}(H \cap K)$.
(In fact we have $\operatorname{cl}(H \cap K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$.)
So that line of approach does not look very fruitful.
So I drew it up as a Venn diagram to get inspiration, and I got:
$$H \cap \operatorname{cl}(K) = H \cap (K \cup \delta K)$$
where $\delta K$ is the boundary of $K$.
This leads us to:
$$H \cap \operatorname{cl}(K) = (H \cap K) \cup (H \cap \delta K) \subseteq (H \cap K) \cup (H \cap \delta K) \cup (\delta H \cap K)$$
and "all that remains to be done" is prove that $(H \cap \delta K) \cup (\delta H \cap K) \subseteq \delta (H \cap K)$ (because $(H \cap \delta K) \cup (\delta (H \cap K)) = \operatorname{cl}(H \cap K)$)
but it's starting to get messy and I'm unsure of my ground.
Besides, this does not make any use of the fact that $H$ is open.
I'm pretty sure there's an insight needed here which I have not been able to come up with.
 A: Hint: Let $x\in H\cap\overline K$. Since $x\in\overline K$, every neighborhood of $x$ contains a point from $K$. Since $H$ is a neighborhood of $x$ [fill in the gaps] it follows that every neighborhood of $x$ contains a point from $H\cap K$.
A: I did not read your proof since I believe is too short by using the definition of $\text{cl}.$ Let $x\in H\cap \text{cl(K)}$ then $x\in H$ and $x\in\text{cl(K)}.$ We need to show that $x\in\text{cl}(H\cap K),$ that is, for any open set $G$ and $x\in G$ we $$G\cap(H\cap K)\neq\emptyset$$
Notice that $G\cap H$ open set and $x\in G\cap H$. But $x\in\text{cl}(K)$. So, we have $(G\cap H)\cap K\neq\emptyset.$ Hence, $H\cap\text{cl}(K)\subset \text{cl}(H\cap K)$
A: $\newcommand{\cl}[1]{\overline{#1}}$
Let $x \in H \cap \cl{K}$. $x \in H$ and $x \in \cl{K}$. The goal is to show that for every neighborhood $N_1$ of $x$, there is an element $y$ of $N_1$ that is also in both $H$ and $K$.
There is a neighborhood $N_2$ of $x$ that is contained in the open set $H$. Let $N_3 = N_1 \cap N_2$. Then $N_3$ is also a neighborhood of $x$ contained in $N_1$. Since $x$ is in $\cl{K}$, by definition of closure, there is an element $y$ of $N_3$ that is also in $K$. From $N_3 \subseteq N_1 \subseteq H$, we know that $y$ is in $H$, by which we showed the existence of such $y \in N_1$ that is in $ H\cap K$. So, we reached the goal.
A: Suppose that $x \in H \cap \operatorname{cl}(K)$.  Then since $x \in \operatorname{cl}(K)$, there exists a net $y : I \to X$ with $y_i \in K$ for every $i \in I$ such that the net $y$ has limit $x$.  Now, since the limit point is in the open set $H$, there exists $i_0 \in I$ such that $x(i) \in H$ whenever $i_0 \le i$.  Therefore, the restriction of $y$ to $\{ i \in I \mid i_0 \le i \}$ is a net with limit $x$ where every element of the net is in $H \cap K$.  This implies that $x \in \operatorname{cl}(H \cap K)$.

If you are not familiar with nets, then you can think of the case where the topological space $X$ is first countable, so that closures are determined by limits of sequences, and then $y : \mathbb{N} \to X$ can be taken to be a sequence instead of a more general net.  The proof then says that for some $i_0$, the sequence $y_{i_0}, y_{i_0+1}, y_{i_0+2}, \ldots$ lies entirely within $H \cap K$ and still has limit $x$.
