Prove that $A_n \cap B_n \rightarrow A \cap B$

The answer to your question comes trivially from the following lemma.

Lemma. $D_n\rightarrow D \Leftrightarrow 1_{D_n}\rightarrow 1_D$

Proof. Let us first show $\Rightarrow$.

Assume that $x\in D$. We need to show that $1_{D_n}(x)\rightarrow 1$. Note that $D=\bigcup_k \bigcap_{n\geq k} D_n$ and therefore, $x\in \bigcup_k \bigcap_{n\geq k} D_n$, i.e., $x\in D_n$ eventually. Therefore, $1_{D_n}(x)=1$ eventually and $1_{D_n}(x)\rightarrow 1$ holds.

Now, assume $x\notin D$. We need to show that $1_{D_n}(x)\rightarrow 0$. Note that $D_n^{c}\rightarrow D^{c}$, where $D^c$ is the complement of $D$. From the previous paragraph we conclude that $1_{D_n^c}(x)\rightarrow 1$ and thus, $1_{D_n}(x)\rightarrow 0$.

We now show $\Leftarrow$.

Let $x\in D$. We have that $1_{D_n}(x)\rightarrow 1$. In particular, $x\in D_n$ eventually. In other words, $D\subseteq \bigcup_{k} \bigcap_{n\geq k} D_n$. Now, we will show that $\bigcap_{k} \bigcup_{n\geq k} D_n\subseteq D$. Consider $x\in D_n$ infinitely often. Then, if $x\notin D$ we would have $1_{D_n}(x)=1$ infinitely often while $1_{D}(x)=0$ which contradicts the fact that $1_{D_n}(x)\rightarrow 1_D(x)$ for all $x$. We then conclude that $x\in D$ or in other words,

$\limsup_n D_n:=\bigcap_{k} \bigcup_{n\geq k} D_n\subseteq D \subseteq \bigcup_{k} \bigcap_{n\geq k} D_n=:\liminf_n D_n$.

This implies $D=\liminf_n D_n=\limsup_n D_n$, i.e., $D_n\rightarrow D$.