= Solution
For every $k\in K$, the <Hausdorff condition> gives <disjoint> <open neighbourhoods> $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an <open cover> of $K$, so choose a <finite subcover> $U_{k_1},\ldots,U_{k_m}$. Put
$$
U=\bigcup_{j=1}^mU_{k_j},\qquad V=\bigcap_{j=1}^mV_{k_j}.
$$
Then $U,V$ are <open sets>, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. Thus
$$
\boxed{K\subseteq U,\qquad x\in V,\qquad U\cap V=\varnothing.}
$$
The finiteness matters: a finite <intersection> of <open sets> is <open>. This is <separation of a point and a compact set>. If $K=\varnothing$, use $U=\varnothing$, $V=X$.
Every $x\in X\setminus K$ consequently has an <open neighbourhood> $V$ contained in $X\setminus K$. The <complement> is <open>, so \b[a <compact subset of a Hausdorff space> is <closed>].
\b[The conclusion can fail without the <Hausdorff condition>.] Take the two-point <set> $X=\{0,1\}$ with the <indiscrete topology> $\{\varnothing,X\}$. The <singleton> $K=\{0\}$ is <compact>, since one member of any <open cover> suffices. It is not <closed>, because its <complement> $\{1\}$ is not <open>. The two points cannot have <disjoint> <open neighbourhoods>, so this example is indeed not <Hausdorff>.
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