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!
Let $(y_k)$ be an arbitrary sequence of terms from $\{x_i\}_{i=1}^{\infty}\cup \{x\}$, and let $d$ be the metric for the metric space. Since $x$ is a limit point, for each $n=1,2,3,\ldots$, there are an infinite number of values of $k$ such that $d(x,y_k)<1/n$. Let $k_1$ be the smallest $k$ such that $d(x,y_{k_1})<1$. For each $i$, define $k_i$ inductively to be the smallest integer such that $k_i>k_{i-1}$ and $d(x,y_{k_i})<1/i$. The subsequence $(y_{k_i})$ converges to $x$ and is a convergent subsequence of $(y_k)$. Thus, every sequence has a convergent subsequence, as claimed.