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Let $A=\{h\} \cup\{h_n\}$ where $h_n \in H$ is a sequence such that $h_n \to h$.

Is $A$ a compact subset of $H$?

I think so, since every subsequence of $A$ has the limit $h$ which is in $A$.

Similarly, $A\backslash \{h\}$ is not a compact subset of $H$.

Is this correct?

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    FYI: just a little mention that your argument only works becasue a Hilbert space is, in particular, a metric space and sequential compactness conicide with compactness – 2017-02-10

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Yes. Another way of seeing it: given an open cover of $A$, take one open that covers $h$, and then only finitely many $h_n$ will be outside and can be covered by finitely many open sets in the cover.

The case about $A\setminus\{h\}$ is slightly different. It could be compact if all $h_n$ become equal eventually.

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    Thanks. If $h_n$ are equal eventually then $h \in A$, no? Since $h_n$ must equal $h$ for large enough $n$. – 2017-02-10
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    I'm thinking of a sequence $h_1,h_2,h_2,h_2,\ldots$. Then $A\setminus\{h\}=\{h_1\}$ is compact. – 2017-02-10