Finite union of compact Hausdorff spaces Is it true that the union of two compact Hausdorff subspaces of an arbitrary space is Hausdorff? What if the ambient space is weakly Hausdorff, that is, the image of any continuous map from a compact Hausdorff space to the given space is closed? (Then it is in particular, a $T_1$-space.)
 A: No, I think the following is a counterexample: let $X = \Bbb N \cup \{-\infty, +\infty\}$ and define the topology on $X$ by declaring all subsets of $\Bbb N$ open, and a subset $A \cup \{-\infty\}$ of $X$ is an open neighbourhood of $-\infty$ iff $\Bbb N\setminus A$ is finite, and similarly for $+\infty$. Then $X$ is not Hausdorff, (as witnessed by $-\infty, +\infty$) despite being the union of the compact Hausdorff subspaces $\Bbb N \cup \{+\infty\}$ and $\Bbb N \cup \{-\infty\}$ (which also have a non-compact intersection BTW).
This $X$ is not weakly Hausdorff as the embedding $N \cup \{\infty\}$ into $X$ does not have a closed image in $X$.
The other answer (by user10354138) shows why that is. The argument only holds for finite unions of compact Hausdorff subsets.
A: The answer to the first question is no, as seen by, e.g., the indiscrete two-point space having the two single point as trivially compact Hausdorff.
The second answer is yes.
Claim If $K$ is compact Hausdorff, $X$ is weakly Hausdorff, $u\colon K\to X$ is continuous, then $u(K)$ is compact Hausdorff in the subspace topology inherited from $X$.
Proof (gist): Let $L=u(K)$ and so $u\colon K\to L$ is closed since any closed subset $E$ of $K$ is compact Hausdorff so $u(E)$ is closed in $X$, hence in $L$ too.  If $a,b\in L$ with $a\neq b$ then $u^{-1}(a),u^{-1}(b)$ are disjoint closed subsets of $K$.  Since compact Hausdorff implies $T_4$, you can find disjoint $U,V$ in $K$ containing $u^{-1}(a)$ and $u^{-1}(b)$ respectively.  Then $U'=\{x\in X\mid u^{-1}(x)\subseteq U\}$ and similarly $V'$ shows $L$ is Hausdorff.$\quad\square$
Apply this to $i_1\cup i_2\colon K_1\amalg K_2\to X$ and you have the union.
