Proof that a limit point compact metric space is compact. If $(X,d)$ is limit point compact, show it is compact.
I have found multiple proofs of this that first show that limit point compact implies sequential compact, which in turn implies compactness. I was wondering if there is a direct proof showing that limit point compactness implies compactness in a metric space.

Pedantic comments are in parentheses.
WLOG, take $C = \{1\} \cup \{1 + \frac{1}{n}\}$. Let $\scr{U}$ be an open cover of $C$. Then $\exists U_d \in \scr{U}$ such that $1 \in U_d$.
Since $U_d$ is open then $1$ is an interior point of $U_d$. Therefore $\exists \epsilon > 0$ such that $N_{\epsilon}(1) \in U_d$.
Note that $N_{\epsilon}(1) = (1 - \epsilon, 1 + \epsilon)$. Now let $N \in \mathbb{N} : N \hspace{3 pt}\bar> \frac{1}{\epsilon}$. Similarly, let $n \in \mathbb{N} : n \overline\geq N$. Then $n > \frac{1}{\epsilon}$ by $\bar ~$ and $\overline ~$.
Therefore $(1 + \frac{1}{n}) - 1 = \frac{1}{n} < \epsilon$ so $1 + \frac{1}{n} \in N_{\epsilon}(1)$ for any $n > N$. (Here the definition of a limit point was used where we removed $\{1\}$ from $C$ and showed it lied in its closure.)
Finally, $$\forall 1 \leq i \leq N - 1\left[\exists U_i \in U : 1 + \frac{1}{i} \in U_i\right]$$ since $C \subset \scr{U}$. (Here we are simply taking the neighborhoods of the finite number of points outside $N_{\epsilon}(1)$. Now we can take $\bigcup_{i = 1}^N U_i$ where $U_N = U_d$. This is the finite sub cover we are looking for (and we shall term this property compactness!!)
Generally, however, we can say,

Let $X$ be a metric in which every infinite subset $A \subset X$ has a limit point. Let $\delta > 0$. Then if $\{x_i\}$ is a sequence in $X$, it converges to some point $x$ such that $d(x_n, x) < \epsilon$ for all $n \leq i$.
Consider $N_{\epsilon}(x)$.  As shown at the top of this answer, $N_{\epsilon}(x)$ is expressible as a union of open balls with rational radius because it contains a countably dense subset. Thus $\bigcup_{i = 1}^n U_i = A$ where $U_n \supset N_{\epsilon}(x)$ and $U_i$ have center $x_i$.

An alternate answer can be found here #24 http://math.mit.edu/~rbm/18.100B/HW4-solved.pdf
