$X$ is a topological space, $C\subset X$. Show that if $X$ is $T_4$ and $C$ is a closed set, then $C$ is also $T_4$.
All I know is that for $X$ to be $T_4$ it has to be also $T_1$ and $\forall_{C,E\subset X;\ C\cap E=\varnothing} \exists_{U,V\subset X}:C\subset U,E\subset V: U\cap V=\varnothing$. Because $C$ is a closed set then it is $T_1$ (at least that's the way I see it, I don't know whether I'm right). My question is how do I prove that $C$ is also $T_4$?
To show that $C$ is $T_1$, let $x,y\in C$ be two distinct points. Since $X$ is $T_1$, there exists an open neighborhood $U$ of $x$ in $X$ such that $y\notin U$. Clearly, we have $x\in U\cap C$ and $y\notin U\cap C$. In particular, $U\cap C$ is open in $C$, so it is an open neighborhood of $x$ in $C$ excluding $y$. This shows that $C$ is $T_1$.
To show $C$ is normal, let $A$ and $B$ be two disjoint closed subsets of $C$. Since $C$ itself is closed in $X$, both $A$ and $B$ are also closed in $X$. Now because $X$ is normal, there are two disjoint open sets $U$ and $V$ in $X$ such that $A\subseteq U$ and $B\subseteq V$. As we can see, $U\cap C$ and $V\cap C$ are open in $C$, where $$A\subseteq U\cap C,\quad B\subseteq V\cap C,\quad(U\cap C)\cap(V\cap C)=\emptyset.$$ Therefore, $C$ is normal.