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.
 A: So, I think I have an idea and would like to run it past someone to see if it holds. I have already proved that if $(X,d)$ is metric and limit point compact then: 


*

*Every open cover of $X$ has a Lebesgue number ($\delta$) and

*For every $\epsilon>0$ there is a finite cover of $X$ by open $\epsilon$-balls (to be notated as $B_\epsilon(x)$, where $x\in X$)


Proof:
Let $A$ be an arbitrary indexing set and $\{U_\alpha\}_{\alpha\in A}$ be an open cover of $X$. Now, $\exists\delta>0$ so that if $D\subseteq X$ such that diam($D$)$<\delta$, then $\exists U_{\alpha_D}$ such that $D\subseteq U_\alpha$. Also, $\exists \{x_1,\dotsc,x_k\}\subseteq X$ such that $X=\bigcup_{i=1}^k B_{\delta/3}(x_i)$. Both of the facts follow from the two statements above.
Now, $\forall 1\leq i\leq k$, since diam($B_{\delta/3}(x_i)$)$<\delta$, $\exists U_{\alpha_i}$ such that $B_{\delta/3}\subseteq U_{\alpha_i}$. So if $x\in X$, then $\exists1\leq n\leq k$ such that $x\in B_{\delta/3}(x_n)\subseteq U_{\alpha_n}$.
So $\{U_{\alpha_i}\}_{i=1}^k$ is a finite subcover of $X$. Therefore $(X,d)$ is compact.
A: 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
A: Suppose $E \subseteq X$ is the limit point compact infinite subset we're talking about. If it's not bounded it's obvious that it has an infinite subset which doesn't have a limit point so it must be bounded.
If it's bounded take $\overline{E}$ so the closure of $E$. This is closed and bounded which is equivalent to compactness in metric spaces. Now if $F \subseteq \overline{E}$ is an infinite subset that has no limit point in $\overline{E}$ then $\forall x \in \overline{E}$ we have a neighborhood such that $\overline{E} \cap N(x)$ either has one element if $x \in F$ or no element if not. But these neighborhoods add an open cover of $\overline{E}$ thus have a finite subcover and at least one of those neighborhoods must have an infinite number of elements which is a contradiction. Note that if $F \subseteq E \subseteq \overline{E}$ then by the previous proof it has a limit point in $\overline{E}$ thus has a limit point but not necessarily in $E$.
