$X$ is normal if for every two open sets $G_1,G_2\subseteq X$ with $G_1\cup G_2=X$, there exists closed sets $F_1\subseteq G_1$ and $F_2\subseteq G_2$ which also satisfies $F_1\cup F_2=X$.
Where can I find the proof for this theorem? I have found multiple proofs for other theorems about normal spaces, but not this one.
Let $H_1,H_2$ be a disjoint pair of closed subsets of $X$. We seek a disjoint open pair $J_1,J_2$ with $H_1\subseteq J_1$ and $H_2\subseteq J_2.$
Let $G_1=X\setminus H_1$ and $G_2=X\setminus H_2.$
Then $G_1,G_2$ are open. And $G_1\cup G_2=X.$ Because (by contradiction) $$(p\in X\land p\not\in G_1\cup G_2)\implies$$ $$ (\,(p\in X\land p\not \in G_1)\land (p\in X\land p\not\in G_2)\,)\implies$$ $$ (\,(p\in H_1)\land (p\in H_2)\,)\implies$$ $$ p\in H_1\cap H_2$$ but by hypothesis we have $H_1\cap H_2=\emptyset.$
Now, by the hypothesis of the Q, take closed $F_1,F_2$ with $F_1\subseteq G_1$ and $F_2\subseteq G_2$ and $F_1\cup F_2=X.$
Let $J_1=X\setminus F_1$ and $=X\setminus F_2.$ Then $J_1, J_2$ are open . And $$J_1\cap J_2=(X\setminus F_1)\cap (X\setminus F_2)=$$ $$=X\setminus (F_1\cup F_2)=$$ $$=X\setminus (X)=\emptyset.$$
Finally we have $$H_1\cap J_1=H_1\cap (X\setminus F_1)=$$ $$=H_1\setminus F_1\supseteq H_1 \setminus G_1 \,\text {(...because } F_1\subseteq G_1...)$$ $$ = H_1 \setminus (X \setminus H_1)=H_1.$$ So $H_1\subseteq J_1.$Changing the subscript from $1$ to $2$ we also have $H_2\subseteq J_2.$