Limit Point Compact subset of an Hausdorff First Countable space is closed So im trying to prove that if i have $X \subset Z$ , where $X$ is limit point compact and $Z$ is Hausdorff and first countable then $X$ is closed in $Z$. Well since $X$ is limit point compact and $Z$ is hausdorff and first countable, and these properties work well with subspaces, i can conclude that $X$ is sequentially compact, but from here i cant seem to prove that $X$ is closed in $Z$, so any tips would be aprecciated.
 A: Assume $X$ were not closed, and take $z\in {\overline X}\cap(Z\setminus X)$. Take a countable base $\{B_n:n\ge1\}$ at $z$ (with $B_{n+1}\subseteq B_n$). For each $n$ pick $x_n\in X\cap B_n$. Do you see how to get a contradiction? Does the seqience $(x_n:n\ge1)$ have a convergent subsequence in $X$? 
A: Suppose $X \subseteq Z$ is limit point compact, and suppose it is not closed, so there is some $p \in \overline{X}\setminus X$. So in particular $X$ is infinite, as all finite subsets of a Hausdorff space are closed ($T_2 \to T_1$).
$p$ has a countable local base $(B_n)_{n \in \Bbb N}$ by $Z$ being first countable, 
and so we can find a sequence of distinct points $x_n \in X$ such that $x_n \to x$ in $Z$: 
First pick $x_1 \in B_1 \cap X$ (as $p \in B_1$ and $p \in \overline{X}$, this is possible). Then having picked $F_n:=\{x_1, \ldots,x_n\} \subseteq X$ all distinct such that each $x_n \in \bigcap_{i=1}^n B_i$, we pick $x_{n+1} \in \left(\bigcap_{i=1}^{n+1} B_i \setminus F_n\right) \cap X$ again by the left set being an open (open minus closed is open) neighbourhood of $p$ and $p \in \overline{X}$, and the recursion continues. If then $O$ is any open neighbourhood of $p$ for some $N$ we have $B_N \subseteq O$ (we have a local base after all) and then for all $n \ge N$, by construction $x_n \in B_N \subseteq O$ as well, so $x_n \to x$. 
Consider the infinite set $N=\{x_n : n \in \Bbb N\}$. As a subset of $X$ there must be a limit point $q \in X$ of $N$ (note that it has to be in $X$, as $X$ is limit point compact by assumption, so $p \neq q$). There are disjoint neighbourhoods $U_q$ and $U_p$ in $Z$ resp. by Hausdorffness and by convergence there is some $n_0$ such that all $x_n$ with $n\ge n_0$ is in $U_p$ and so misses $U_q$. Also, $F=\{x_1, x_2, \ldots, x_{n_0}\} \setminus \{q\}$ is finite and thus closed and so
$(U_q \setminus F)$ is an open neighbourhood of $q$ that is disjoint from $N$, contradicting it would be a limit point of that set. Contradiction. 
And so $X$ must be closed in $Z$. We do need $T_2$ and not just $T_1$ (as we indeed used) or otherwise $Z = \Bbb Z$ in the cofinite topology and $X=\Bbb N$ would have been a counterexample. ($X$ is compact so limit point compact, and a countable cofinite space is $T_1$ and first countable.)
