$X$ is weak Hausdorff space means for any compact Hausdorff space $C$ and $u: C\to X$ continuous, $\text{im } u$ is closed
We know that the property in the title holds for Hausdorff spaces, is this true for weak Hausdorff spaces?
It is noted in nlab without proof:
For the most common purposes for which Hausdorff spaces are used, the assumption of being weakly Hausdorff suffices.
hence the question
Theorem: A topological space $X$ is Hausdorff if and only if every convergent net in $X$ converges to only one point.
Proof: Let $X$ be Hausdorff, $(x_\alpha)_{\alpha\in A}$ a convergent net, and $\xi$ a limit point of $(x_\alpha)$. If $\eta\neq \xi$, then by the Hausdorff property, there are disjoint neighbourhoods $U,V$ of $\xi$ and $\eta$ respectively. Since $x_\alpha \to \xi$, there is an $\alpha_U$ such that $x_\alpha \in U$ for all $\alpha \geqslant \alpha_U$. But then $x_\alpha \notin V$ for any $\alpha \geqslant \alpha_U$, and hence $\eta$ is not a limit point of the net. So the limit point is unique in Hausdorff spaces.
If $X$ is not Hausdorff, there are two points $\xi \neq \eta$ such that every neighbourhood of $\xi$ intersects every neighbourhood of $\eta$. Let $\mathscr{V}(x)$ denote the neighbourhood filter of a point $x\in X$. On $A := \mathscr{V}(\xi) \times \mathscr{V}(\eta)$, define the partial order
$$(U_1,V_1) \leqslant (U_2,V_2) \iff (U_2 \subset U_1 \land V_2\subset V_1).$$
With this partial order, $A$ is a directed set, and we define a net by choosing, for all $(U,V)\in A$ a point $x_{(U,V)} \in U\cap V$ (which is not empty by the assumption). Then the net $(x_{(U,V)})_{(U,V)\in A}$ converges to $\xi$ as well as to $\eta$. Hence there is a convergent net with more than one limit in every non-Hausdorff space.