Compact subspace of a Hausdorff space is closed, and compact Hausdorff implies normal

1. Let \(K\subseteq X\) be compact and \(X\) Hausdorff. Fix \(x\in X\setminus K\). For each \(y\in K\), choose disjoint open neighborhoods \[ x\in U_y,\qquad y\in V_y. \] The sets \(V_y\) cover \(K\), so compactness gives \(y_1,\dots,y_n\) with \(K\subseteq\bigcup_iV_{y_i}\). Then \[ U=\bigcap_{i=1}^nU_{y_i} \] is an open neighborhood of \(x\) disjoint from \(K\). Thus \(X\setminus K\) is open and \(K\) is closed.

Hausdorffness is necessary. In the Sierpiński space \[ X=\{0,1\},\qquad\tau=\{\varnothing,\{1\},X\}, \] the singleton \(\{1\}\) is compact but not closed.

2. Let \(A,B\subseteq X\) 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 \(U_{a,b}\ni a\), \(V_{a,b}\ni b\). For fixed \(a\), compactness of \(B\) gives \(b_1,\dots,b_m\) with \(B\subseteq\bigcup_jV_{a,b_j}\). Set \[ U_a=\bigcap_jU_{a,b_j}, \qquad V_a=\bigcup_jV_{a,b_j}. \] Then \(a\in U_a\), \(B\subseteq V_a\), and \(U_a\cap V_a=\varnothing\). Compactness of \(A\) gives \(a_1,\dots,a_r\) with \(A\subseteq\bigcup_iU_{a_i}\). Finally set \[ U=\bigcup_iU_{a_i}, \qquad V=\bigcap_iV_{a_i}. \] Then \(U,V\) are disjoint open sets containing \(A,B\), respectively. Hence \(X\) is normal.