I want to show that if $(x_n)$ is a convergent sequence in a metric space, $(x_n) \rightarrow x$, then the subset $X = \{x_1,x_2,...\}\cup{x}$ is compact. Does this proof suffice?
Since $(x_n)$ converges, all of its subsequences must also converge. Now to show $X$ is compact, we need to show any sequence in $X$ has a convergent subsequence. So consider $(y_n) \subset X$. If $y_n = x$ for all $n$, we are done. Similarly, if $y_n$ eventually has infinitely many terms equal to $x$, take that subsequence.
So we are left with the case where only finitely many terms from $y_n$ are $x$. Writing $(y_n) = \{y_1,y_2,...\}$, take the subsequence given by deleting terms equal to $x$, so $y_{n_k}=\{y_1,y_2,...\}\backslash \{x\}$ This is now clearly a subsequence of $(x_n)$, so must converge. Thus $X$ is compact.
That seems too straightforward, am I missing something? Thanks!