= Solution
Suppose $K\setminus\{a\}$ were <compact>. By the <compact subset of a Hausdorff space> theorem, it would be <closed> in $X$, hence <closed> in $K$. The <singleton> $\{a\}$ is also <closed> in the <Hausdorff space> $X$: each other point has an <open> <neighbourhood> avoiding $a$.
Consequently both $\{a\}$ and $K\setminus\{a\}$ are <clopen> in $K$. They are disjoint, nonempty, and cover $K$, giving a <separation of a topological space> in contradiction to the <connectedness> of $K$. Thus $K\setminus\{a\}$ is not <compact>.
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