Subspace closed iff compact implies Hausdorff? This site claims without proof that the following are equivalent for a topological space $X$:

*

*$X$ is compact and Hausdorff

*A subspace of $X$ is compact iff it is closed in $X$.

Is this claim correct?  It should be easy to deduce if true. It is quite easy to see this condition implies compact $\mathop T_1$, but I don’t think it would imply Hausdorff. Is there any counterexample or proof?
 A: Here's a counter-example.
Let $X = \hat{\mathbb{Q}}$ be the one-point compactification of $\mathbb{Q}$. It's well-known that such compactification is Hausdorff iff the underlying space is locally compact and Hausdorff, so $X$ is not Hausdorff.
If $Y\subseteq X$ is closed then it's compact as a closed subset of compact space, so we only need to consider if compact subsets are closed.
Let $Y\subseteq X$ be compact. If $Y\subseteq \mathbb{Q}$ then $Y$ is closed in $X$ by definition. Otherwise we can write $Y= Y_0\cup \{\infty\}$ where $Y_0\subseteq \mathbb{Q}$. It suffices to prove that $Y_0$ is closed in $\mathbb{Q}$. From compactness of $Y$ it follows that if $K\subseteq \mathbb{Q}$ is compact then $Y_0\cap K$ is compact. To see this, cover it by any family of open subsets of $\mathbb{Q}$ and add $(\mathbb{Q}\setminus K)\cup \{\infty\}$ to it. From compactness of $Y_0$ we can extract an open subcover, and we can remove the additional set as it doesn't intersect $Y_0\cap K$. Take a sequence $y_n\in Y_0$ convergent in $\mathbb{Q}$ to $y$, and consider $K = \{y_n: n\in\mathbb{N}\}\cup \{y\}$, then from compactness of $Y_0\cap K$ we have $y\in Y_0$. This proves that $Y_0$ is closed in $\mathbb{Q}$.
Note that I haven't used any properties other than that $\mathbb{Q}$ is a metric space which is not locally compact. This gives us a whole family of counter-examples.
