= Solution
A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.
Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the <open set> $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a <compact space> is compact.
Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint <open sets> $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. Put
$$
U_a=\bigcap_jU_{a,b_j},
\qquad
V_a=\bigcup_jV_{a,b_j}.
$$
Then $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Therefore
$$
U=\bigcup_iU_{a_i},
\qquad
V=\bigcap_iV_{a_i}
$$
are disjoint open neighbourhoods of $A$ and $B$. This proves the <normality of a compact Hausdorff space>.
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