If $H$ and $K$ are abelian subgroups of a group $G$, then $H\cap K$ is a normal subgroup of $\left\langle H\cup K\right\rangle$.
I proved $H\cap K$ is a subgroup and need to prove it is normal subgroup of $\left\langle H\cup K\right\rangle$. But isn't it obvious that $H\cap K$ is normal? All the elements in $H\cap K$ communicates with all the elements in $H$ and $K$, hence $H\cap K$ is normal. Am I missing something?
On
Suppose $L$ is a subgroup of $\langle X\rangle$, where $X$ is a subset of a group $G$. In order to show that $L$ is normal in $\langle X\rangle$, it is sufficient to prove that, for every $z\in L$ and $x\in X$, $xzx^{-1}\in L$.
This is because, defining $\varphi_x\colon G\to G$ by $\varphi_x(g)=xgx^{-1}$, we have $\varphi_{xy}=\varphi_x\circ\varphi_y$ and $\varphi_x^{-1}=\varphi_{x^{-1}}$.
In your case $X=H\cup K$; if $x\in H$, then $xzx^{-1}=z$, as this operation is inside $H$. Similarly if $x\in K$.
Yes, every element of $H\cap K$ commutes with every element of $H$ and with every element of $K$. In my opinion, you should add a proof of that fact that it follows from this that every element of $H\cap K$ commutes with every element of $\langle H\cup K\rangle$. You can say, for instance, that every element of $\langle H\cup K\rangle$ can be written as a product of elements of $H\cup K$ and then use that fact.