Could someone please verify my following proof?
If $H_{1}\unlhd G$ and $H_{2}\unlhd G$ with $G$ being a group, then $H_{1}\cap H_{2}\unlhd G$.
Proof: Let $H_{1}\unlhd G$ and $H_{2}\unlhd G$. Then if $ghg^{-1}\in H_{1}\cap H_{2}$ for any $g\in G$ and $h\in H_{1}\cap H_{2}$, then $gHg^{-1}\subset H_{1}\cap H_{2}$, which implies that $H_{1}\cap H_{2}\unlhd G$.
Let $g\in G$ and $h\in H_{1} \cap H_{2}$. Since $h\in H_{1}\cap H_{2}$, $h\in H_{1}$. Since $H_{1}$ is normal, $ghg^{-1}\in H_{1}$. Since $h\in H_{1}\cap H_{2}$, $h\in H_{2}$. Since $H_{2}$ is normal, $ghg^{-1}\in H_{2}$. Since $ghg^{-1}\in H_{1}\cap H_{2}$, $H_{1}\cap H_{2}\unlhd G$. $\square$