These are the completely ultrametrizable spaces.
Recall that a d:E2→[0,∞) is an ultrametric if
- d(x,y) = 0 ↔ x = y
- d(x,y) = d(y,x)
- d(x,z) ≤ max(d(x,y),d(y,z))
As usual, (E,d) is a complete ultrametric space if every Cauchy sequence converges.
Suppose E∞ is the inverse limit of the sequence En of discrete spaces, with fn:E∞→En being the limit maps. Then
d∞(x,y) = inf { 2-n : fn(x) = fn(y) }
is a complete ultrametric on E∞, which is compatible with the inverse limit topology.
Conversely, given a complete ultrametric space (E,d), the relation x ∼n y defined by d(x,y) ≤ 2-n is an equivalence relation. Let En be the quotient E/∼n, with the discrete topology. These spaces have obvious commuting maps between them, let E∞ be the inverse limit of this system. The map which sends each point of E to the sequence of its ∼n equivalence classes is a continuous map f:E→E∞. Because E is complete, this map f is a bijection. Moreover, a simple computation shows that this bijection is in fact a homeomorphism. Indeed, with d∞ defined as above, we have
d∞(f(x),f(y)) ≥ d(x,y) ≥ d∞(f(x),f(y))/2.
As Pete Clark pointed out in the comments, the above is an incomplete answer since the question does not assume that the inverse system is countable. However, the general case does admit a similar characterization in terms of uniformities. For the purposes of this answer, let us say that an ultrauniformity is a unformity with a fundamental system of entourages which consists of open (hence clopen) equivalence relations. The spaces in question are precisely the complete Hausdorff ultrauniform spaces.
Suppose E is the inverse limit of the discrete spaces Ei with limit maps fi:E→Ei. Without loss of generality, this is a directed system. Then the sets Ui = {(x,y): fi(x) = fi(y)} form a fundamental system of entourages for the topology on E, each of which is a clopen equivalence relation on E. The universal property of inverse limits guarantees that E is complete and Hausdorff. Indeed, every Cauchy filter on E defines a compatible sequence of points in the spaces Ei, which is the unique limit of this filter.
Conversely, suppose E is a complete Hausdorff ultrauniform space. If U is a fundamental entourage (so U is a clopen equivalence relation on E) then the quotient space E/U is a discrete space since the diagonal is clopen. In fact, E is the inverse limit of this directed system of quotients. It is a good exercise (for Pete's students) to show that completeness and Hausdorffness of E ensure that E satisfies the universal property of inverse limits.
Expanding on my comment, if there are measurable cardinals then it follows from the results of
A. Przeździecki, Measurable cardinals and fundamental groups of compact spaces.
Fund. Math. 192 (2006), no. 1, 87–92.
that there are spaces not weakly equivalent to any compact Hausdorff space, as Przeździecki proves that (if and only if there is a measurable cardinal) there are groups $G$ that are not the fundamental group of any compact Hausdorff space, and so the classifying space of such a group is a counterexample.
He also proves that every group of non-measurable cardinality is the fundamental group of a path-connected compact space, answering a question of Keesling and Rudyak who had earlier proved this with "connected" in place of "path-connected" in
J.E. Keesling and Y.B. Rudyak, On fundamental groups of compact Hausdorff spaces.
Proc. Amer. Math. Soc. 135 (2007), no. 8, 2629–2631.
Best Answer
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 ... ???