Relationship between weak Hausdorff and US properties A space is called weak Hausdorff if the image of every continuous map from a compact Hausdorff space into the space is closed.
I am trying to see where the weak Hausdorff property fits in the following hierarchy of properties between $T_1$ and $T_2$:
$$T_2\implies KC\implies US\implies T_1.$$
Here, a space is called KC if all compact subsets are closed.  And a space $X$ is US ("unique sequential limits") if every convergent sequence of elements of $X$ has a unique limit in $X$.
Clearly, KC implies weak Hausdorff.  I think the following should be true:

Every weak Hausdorff space is US.

(1) Can anyone check that the proof below is correct?
Given that, the chain of implications becomes:
$$T_2\implies KC\implies \text{weak Hausdorff}\implies US\implies T_1.$$
The implications in the chain cannot be reversed.  An example of weak Hausdorff space that is not KC is $\Bbb Q^*\times\Bbb Q^*$, where $\Bbb Q^*$ is the one-point compactification of $\Bbb Q$, as explained here.
(2) What would be an example of a US space that is not weak Hausdorff?

Proof that weak Hausdorff implies US:
Let $X$ be a weak Hausdorff space.  Let $(x_n)$ be a sequence in $X$ that converges to $x\in X$.  We have to show that the limit is unique.
The space $Y=\{0\}\cup\{1,\frac{1}{2},\frac{1}{3},...\}$ with the subspace topology induced from $\Bbb R$ is compact Hausdorff.  The map $f:Y\to X$ defined by $f(\frac{1}{n})=x_n$ and $f(0)=x$ is continuous at every point $\frac{1}{n}$ as such points are isolated in $Y$, and is continuous at $0$ as the sequence converges to $x$.  So the image $f(Y)=\{x\}\cup\{x_1, x_2,...\}$ is closed in $X$.  Thus any limit of $(x_n)$ in $X$ must belong to $f(Y)$.  Furthermore, as shown in Lemma 1 here, $f(Y)$ is Hausdorff.  So the limit of $(x_n)$ is unique.
 A: I think that your proof that wH implies US is correct.
Maybe at the end you don't need the quoted lemma 1 at all: if $x_n \to x'$ where $x' \in X$ with $x' \neq x$, too, then indeed $x' \in \overline{f[Y]}=f[Y]$ but clearly $x' \notin f[Y]$...
For the example, you need to go into the obscure spaces by van Douwen and some others that are US but not KC, I think. They're pretty rare, because there isn't much interest in these weak separation properties anyway.
A: Here are some examples of US spaces that are not weak Hausdorff.  They were provided by Paul Fabel in this math overflow post.  In each case, the space is not KC, since it contains a compact subspace that is not closed; but the subspace is also Hausdorff, showing that the space is not even weak Hausdorff.
Example 1: Let $\omega_1$ be the first uncountable ordinal with the usual order topology.  Take the one-point compactification $\omega_1+1=[0,\omega_1]$ and then split the maximum point into two points, similar to the contruction of the "line with two origins".  In other words, take $X=[0,\omega_1]\cup\{\omega'_1\}$ where sets of the form $(\beta,\omega_1)\cup\{\omega'_1\}$ with $\beta<\omega_1$ form a nbhd basis at $\omega'_1$.
The subset $[0,\omega_1]$ is compact Hausdorff, but not closed in $X$, so $X$ is not weak Hausdorff.  It is also not too difficult to check that $X$ is US.  (Sequences in $[0,\omega_1)$ cannot converge to $\omega_1$ or to $\omega'_1$; and sequences that take the value $\omega_1$ infinitely many times can only converge to $\omega_1$ by the $T_1$ property, and similarly for $\omega'_1$.)
Example 2: Let $I=[0,1]$ with its usual topology. Take $X=I\cup\{z\}$ for an additional point $z$.  Open subsets of $I$ remain open in $X$, and open nbhds of $z$ are the subsets containing $z$ and whose intersection with $I$ is open dense in $I$ (in other words the complements in $X$ of closed nowhere dense subsets of $I$).
The subspace $I$ is compact Hausdorff, but not closed in $X$; hence $X$ is not weak Hausdorff.  The fact that $X$ is US is shown in detail in this post.
