Is this kind of space metrizable? It has a nice result from Tkachuk V V. Spaces that are projective with respect to classes of mappings[J]. Trans. Moscow Math. Soc, 1988, 50: 139-156.

If the closure of every discrete subset of a space is compact then the whole space is compact.

The proof can be seen here by Brian M. Scott.
Then these questions arise naturally:
Question 1: If the closure of every discrete subset of a space is countably compact, then  is the whole space countably compact?
Question 2: If the closure of every discrete subset of a space is metrizable, then is the whole space metrizable?
 A: $\newcommand{\cl}{\operatorname{cl}}$The first conjecture is true at least for $T_1$ spaces.
If $X$ is $T_1$ and not countably compact, then $X$ has an infinite closed discrete subspace, which is obviously not countably compact. Thus, if every discrete subspace of a $T_1$ space $X$ is countably compact, so is $X$. 
It’s at least consistent that the second conjecture is false. It is consistent that there be a compact Suslin line, i.e., a complete dense linear order $\langle X,\preceq\rangle$ with endpoints such that the order topology on $X$ is ccc but not separable. (E.g., the existence of Suslin line follows from the combinatorial principle $\diamondsuit$, which holds in $\mathsf{V=L}$.) 
Suppose that $F\subseteq X$ is closed, $x\in F$, and $F\cap[x,\to)$ is open in $F$. If $x=\min F$, then $x$ is not a left pseudogap of $F$. (For a definition of left and right pseudogaps see this answer.) Otherwise, $F\cap(\leftarrow,x)$ is a non-empty closed subset of $F$ and therefore of $X$. It follows that $F\cap(\leftarrow,x)$ is compact and has a maximum element $y$. But then $F\cap[x,\to)=\{z\in F:y\prec z\}$ is open in the order topology on $F$, and $x$ is not a left pseudogap of $F$. A similar argument shows that $F$ has no right pseudogaps and hence that the subspace topology on $F$ is identical to the order topology, so that $F$ with its subspace topology is a LOTS.
Let $D\subseteq X$ be discrete. The spread of any LOTS is equal to its cellularity, so $D$ is countable. Let $Y=\cl_XD$; then $Y$ is a separable, compact LOTS. Let $$J=\{x\in Y:x\text{ has an immediate successor in }Y\}\;;$$ since $Y$ is a LOTS, $w(Y)=c(Y)+|J|=\omega+|J|$. For $x\in J$ let $x^+$ be the immediate successor of $x$ in $Y$; then $\{(x,x^+):x\in J\}$ is a pairwise disjoint family of non-empty open intervals in $X$, so $|J|\le\omega$, and $w(Y)=\omega$. It now follows from the Uryson metrization theorem that $Y$ is metrizable and hence that every discrete subset of $X$ has metrizable closure.
A: Q1: (Assuming $T_1$-ness) If $X$ were not countably compact then there is a countable subset $A$ without a limit point. But this $A$ is then closed and discrete, so $A$ is the closure of a discrete set, hence countably compact, which it obviously is not. So the question seems trivial for the countably compact case. 
Q2: There is a consistent counterexample at least (I studied that space in the past) in this paper (not free) by Kenneth Kunen. But I think there should be a much easier example.
A: For question 2, I think that any non-metrizable compact Hausdorff space works, since the only discrete spaces are finite.
