This is Exercise 3 c) from Section 28 of Munkres – Topology. I had thought the answer was no initially, but the example I came up with was not Hausdorff. At this point, I'm just not sure. Here's what I've done so far:
Suppose $X$ is not closed. Then $X$ has a limit point $x\in Z\setminus X$. Let $U_0$ be neighborhood of $x$, then there is an $x_1\in U_0\cap X$. Because, $Z$ is Hausdorff, $U_0$ is Hausdorff. So, we can recursively define for each $n$, distinct $x_n\in U_{n-1}$ and disjoint neighborhoods $V_n,U_n\subseteq U_{n-1}$ of $x_n,z$ respectively. Because, $V_n\subseteq U_{n-1}\setminus U_n$, $V_n\cap V_m=\emptyset$ whenever $n\ne m$. Therefore, each $x_n$ is an isolated point.
My problem now is that I can't claim that $x$ is a limit point of $\{x_n\}$. I may well not be if $Z$ is not first countable.
Best Answer
And in fact the first counterexample that comes to mind works precisely because the point $x_0$ is not a point of first countability. I don’t remember whether Munkres uses the standard name $\omega_1$ or calls it $\Omega$, but you want the space $Z$ to be the linearly ordered space of ordinals less than or equal to the first uncountable ordinal, $[0,\omega_1]$ or $[0,\Omega]$, depending on your notation. $Z$ is even a compact Hausdorff space. Let $X=[0,\omega_1)$ (or $[0,\Omega)$); it’s a standard example of a limit point space that is not compact, and it’s clearly not closed in $Z$.