I have a false proof that all Hausdorff spaces are normal; I know the correct theorem is
All compact Hausdorff spaces are normal.
I can't figure out what I'm missing in my proof, I seem to be able to get away it without using normality. (1) I define a topological space $X$ to be Hausdorff iff for all non-identical points $p, q \in X$ , we have two disjoint open neighbourhoods of $p, q$. Formally, we have two open sets $O_p, O_q$ such that $p \in O_p$, $q \in O_q$, and $O_p \cap O_q = \emptyset$. (2) A topological space is normal iff for all disjoint closeds set $C, D \subseteq X$, we have two disjoint open neighbourhoods of $C, D$. Formally, we have two open sets $O_C, O_D$ such that $C \subseteq O_C, D \subseteq O_D$ and $O_C \cap O_D = \emptyset$.
Let $X$ be Hausdorff. I will show that it is normal. Let $C, D$ be two disjoint closed sets (so $C \cap D = \emptyset$). Pick open covers $O_C, O_D$ of $C$ and $D$ respectively. Now restrict the open cover $O_C$ to $D^c$. This is possible since $D^c$ is open, and this new restricted cover $O'_C$ continues to cover $C$, but does not cover $D$. Similarly, define $O'_D$ by restricting $O_D$ to $C^c$. Formally, $O'_C \equiv O_C \cap D^c = \{ O \cap D^c : O \in O_C \}$ and similarly $O'_D \equiv O_D \cap C^c = \{ O \cap C^c : O \in O_D \}$.
We have that $O'_C \cap O'_D = \emptyset$, and $C \subset \bigcup O'_C$, and $D \subset \bigcup O'_D$. We have thus separated $C$ and $D$ by opens.
Where am I going wrong, since I don't use compactness anywhere?