Let $X$ be a $T_4$-space and $M \subset X$, then $M$ is a subspace that is also $T_4$.
Proof: If $A,B$ are closed in $M$, then $A=W_A \cap M$ and $B = W_B \cap M$ for some closed sets $W_A,W_B$. Since $X$ is $T_4$, we can separate them by open sets $W_A\subset U_A$, $W_B\subset U_B$ and therefore $A,B$ are separated by $U_A \cap M, U_B \cap M$.