I've looked on the web and haven't found a simple example.

7$\begingroup$ From Wikipedia: en.wikipedia.org/wiki/Weak_Hausdorff_space A space $X$ is weak Hausdorff if whenever $Y$ is compact Hausdorff and $f:Y\rightarrow X$ is continuous, then $f(Y)$ is closed in $X$. $\endgroup$ – Matthew Daws Feb 14 '12 at 11:16

$\begingroup$ Welcome to MathOverflow, Prof. Solovay. Gerhard "And Happy Valentines Day, Too" Paseman, 2012.02.14 $\endgroup$ – Gerhard Paseman Feb 14 '12 at 20:15
The onepoint compactification of $\mathbb{Q}$ has the property that every compact subset is closed. So it is certainly a weak Hausdorff space. But it isn't Hausdorff, as $\mathbb{Q}$ isn't locally compact.
Addendum
Another example is the cocountable topology on an uncountable set. No two points have disjoint neighbourhoods, and the only compact subsets are the finite subsets.

1$\begingroup$ I don't know what you mean by "the one point compactification of Q". Of course, I am familiar with the one point compactification of a locally compact space. What is the topology of this space. What are two points that don't have disjoint neighborhoods? $\endgroup$ – Bob Solovay Feb 15 '12 at 2:21

2$\begingroup$ @BobSolovay: I was not sure of this myself, but consulting Kelley's book tells me that for any top. space X the 1point compactification X* has as its open sets all the open subsets of X, together with all subsets of X* whose complements are closed compact subsets of X. (TBC) $\endgroup$ – Yemon Choi Feb 15 '12 at 3:46

$\begingroup$ Thus, if I have understood things correctly: when we give Q its subspace topology from R (thus not locally compact) we find that the open neighbourhoods of the point at infinity correspond to cofinite subsets of Q, and hence they meet every nonempty open subset of Q. (Open subsets of Q in this topology are either empty or infinite.) $\endgroup$ – Yemon Choi Feb 15 '12 at 3:48

7$\begingroup$ @Bob Solovay: Yemon Choi's definition of the onepoint compactification is correct. @Yemon Choi: But your second comment is wrong, since $\mathbb{Q}$ has infinite compact subsets (e.g., any convergent sequence together with its limit forms a compact set). But every compact subset of $\mathbb{Q}$ has empty interior, so it's true that every neighbourhood of the point at infinity meets every nonempty open set. $\endgroup$ – Stephen S Feb 15 '12 at 9:05
Steen & Seebach's counterexample #99: Maximal Compact Topology is another example. This is also a KC space (every compact set is closed) but not Hausdorff.