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A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers X

It is known that a Hausdorff space is compact iff it is a $D$ space and countably compact. It is also known that every countably compact is star compact and the converse holds whenever the space is Hausdorff.

Is it true that every $T_1$ is compact iff it is a $D$ space and star compact?

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    What's your definition of star compact? There are two that occur in the literature. – 2017-02-02
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    @henno Brandsma: $X$ is star compact if every open cover $\mathcal{U}$ of $X$ there is a finite set $A$ such that $st(A, \mathcal{U})= X$. – 2017-02-03
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    OK, star finite in other words – 2017-02-03
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    So basically you want a generalisation of Hausdorff to only $T_1$. Where in the proof (which I suppose you know) do you use Hausdorffness? This might give you a clue to look for an example. – 2017-02-03
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    We use Hausdorffness to define countable open cover which doesn't have finite subcover. – 2017-02-03
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    You follow the proof in the van Douwen et al. paper http://ac.els-cdn.com/016686419190077Y/1-s2.0-016686419190077Y-main.pdf?_tid=1f011992-e9d0-11e6-96c2-00000aacb361&acdnat=1486099256_81cb81c2b881d2e37024164ed56d1e65 thms 2.1.4 and 2.1.5. O I presume? – 2017-02-03
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    Yes, using Hausdorffness every two distinct points can be separated by two neigbourhoods. But it does not hold for $T_1$. So in a Hausdorff space, we can construct such open cover . – 2017-02-03

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