Is a inverse limit of compact spaces again compact ?

What does this example do ...

All spaces are on set $\{1,2,\dots\}$. Space $X_n$ has topology that makes $\{1,2,\dots,n\}$ discrete and $\{n+1,\dots\}$ indiscrete. Of course $X_n$ is compact non-Hausdorff. Map $X_{n+1} \to X_n$ by the "identity". Inverse limit is ... ???

Although the question is quite old I just ran into it. You might still want to (if you haven´t done so yet) look at the article:

A.H. Stone, Inverse limits of compact spaces, General Topology and its Applications 10 Issue 2 (1979) pp 203–211 https://doi.org/10.1016/0016-660X(79)90008-4.

It contains some positive results.

Let $X'=\{\frac1n:n\in\mathbb N\}\subset\mathbb R$, with its subspace topology, and let $X=X\cup\{0\_1,0\_2\}$ be the result of adding two distinct limits to the obvious sequence; $X$ is a compact non-Hausdorff space. Let $s:X\to X$ be the map which swaps the two zeroes, which is continuous. Then the equalizer of $s:X\to X$ and $\mathrm{id}\_X:X\to X$ is $X'$, which is not compact. If in 'inverse limit' you include general limits, then the answer is thus no.