Ok, so we want something like $\operatorname{supp} u \subseteq \cap_{N=1}^\infty \cup_{n=N}^\infty \operatorname{supp} u_n$. Here we go:
$(u_n)_{n \in \mathbb{N}}$ converges weakly to $u$, and so does every restricted sequence $(u_n)_{n \geq N}$. $A_N = \{ f \in L^2; \operatorname{supp} f \subseteq \cup_{n=N}^\infty \operatorname{supp}u_n\}$ is a closed subspace of $L^2$, and closed implies weakly closed. Therefore $u$ needs to be in $A_N$ for each $N$. In particular,
$$ \operatorname{supp} u \subseteq \cap_{N=1}^\infty \cup_{n=N}^\infty \operatorname{supp} u_n.$$
Edit: A sketch of why I think the statement will be wrong if we replace the lim sup by lim inf:
Consider $L^2[0,1]$ and let the functions $f_{k,n}$ be defined by
$$ f_{k,m} = 1_{\big [ \frac{k}{m},\frac{k+1}{m}\big ]}$$
for $m \in \mathbb{N}$ and $k\in \mathbb{N}, 0 \leq k < m$. Enumerate them somehow as $(f_n)_{n \in \mathbb{N}}$. Then $f_n \to 0$ in norm, hence $f_n \to 0$ weakly. Consider
$$ u_n = 1 - f_n.$$
Then $u_n \to 1$ weakly. But on the other hand, for each $x \in [0,1]$ there are infinitely many $n \in \mathbb{N}$ such that there is an open interval $I(n)$ containing $x$ and $\operatorname{supp} u_n \cap I(n) = \emptyset$. Therefore $\lim \inf \operatorname{supp} u_n = \emptyset$ and $\operatorname{supp} u = [0,1]$.