Convergent Sequence + Limit is Compact using Sequential Compactness Proposition: Let $(X,d)$ be a metric space and $\lbrace x_n \rbrace_{n=1}^\infty \subset X$ be a convergent sequence with $x_n \rightarrow x_0, n \rightarrow \infty$. Show that $K = \lbrace x_n \mid n \in \mathbb{N} \cup \lbrace 0 \rbrace \rbrace$ is a compact set.
Question: It is easy to see how to do this with open covers via the standard definition of compactness. What I am wondering is how one would prove this theorem using sequential compactness (since it is equivalent to regular compactness for metric spaces), if it is even possible to do so.
 A: Suppose $\{z_n\}$ is a sequence in $K$.
If the sequence visits infinitely often a point, then we get a convergent subsequence. Thus we can assume the sequence only visits each point a finite number of times (in particular, $K$ is infinite).
Choose $k(1)$ so that $z_{k(1)}$ is in $K'=\{x_n:n>0\}$ and let $p(1)$ be the least integer such that $z_{k(1)}=x_{p(1)}$.
Suppose we have selected $k(n)$ and $p(n)$. Choose $k(n+1)$ to be the least integer $m$ such that $m>k(r)$, $z_{m}\in K'$ and $z_m=x_r$ for some $r>p(n)$; $p(n+1)$ will be the least such $r$.
Note that $z_{k(n)}=x_{p(n)}$, and that $\{x_{p(n)}\}$ is a subsequence of the convergent sequence we started with.
A: I will rewrite my comment to egreg's solution as an answer to see if it raises some objections.
Suppose $(z_n)$ is a sequence in $K$. We only have two cases.

*

*$(z_n)$ attains some value of $K$ infinitely many times, in which case it obviously has a convergent (constant) subsequence that converges in $K$, or

*$(z_n)$ attains all values of $K$ a finite number of times. In this case $(z_n) \to x_0$.

To show 2., for $i\geq 1$, consider the finite set $$K_i = \{x_1,x_2,\dots,x_i\}.$$ Since none of the values of $K_i$ are reached infinitely many times by $(z_n)$, then there exists $N_i$ such that $z_n\not \in K_i$, for $n>N_i$.
Fix $\varepsilon > 0$. By convergence of $(x_n)$, there exists  $M_\varepsilon$ such that $d(x_n,x_0)< \varepsilon$, for all $n>M_\varepsilon$. Now, for $n>N_{M_\varepsilon}$ we have $d(z_n,x_0) < \varepsilon$, thus proving the convergence of $(z_n)$ to $x_0$. $\blacksquare$
