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I know that this is true for Hausdorff spaces and metric spaces, which are Hausdorff spaces, but I can’t prove it for second countable spaces. Is it even true? Thanks!

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No, it's not true. For the easiest example take a two point space where one of the points is open and the other is not. The point that is open is not closed, but it is compact because it is finite. Clearly every finite space is second countable.

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    $\begingroup$ Or for an even easier example take the indiscrete topology. $\endgroup$ – Eric Wofsey Jan 20 at 21:51
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    $\begingroup$ @EricWofsey True, that would also work. $\endgroup$ – Matt Samuel Jan 20 at 21:52
  • $\begingroup$ Oh yes, you’re right. Thanks a lot $\endgroup$ – Brian Mac Guire Jan 20 at 22:37
  • $\begingroup$ Then, how can I prove, for second countable spaces, that compact subset iff sequentially compact? Thanks! $\endgroup$ – Brian Mac Guire Jan 20 at 22:41
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$X=\Bbb N$ in the co-finite topology is $T_1$ but not Hausdorff and trivially second countable as there are only countably many open sets, and every subset of it is compact, but only the finite ones and $X$ are closed...

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