Let $u_n \to u$ strongly in $C([0,T];H^{-1}(\Omega))$ and suppose that $u_n$ is uniformly bounded in $L^\infty((0,T)\times\Omega)$. Then $$u_n(t) \to u(t)$$ weakly in $L^1(\Omega)$ for each $t$?
Can someone show me what result this is? How it is gotten?
I think you need $\Omega$ of bounded measure (e.g. bounded).
Fix $t\in[0,T]$. You have $u_n(t)\to u(t)$ in $H^{-1}$. Furthermore, $\sup_n \|u_n(t)\|_\infty=M<\infty$ by assumption.
Therefore your sequence $u_n$ is uniformly integrable, which means, that for every $\epsilon>0$, there exists $\delta>0$ such that for all measurable set $A\subset\Omega$, if $|A|<\delta$ then $\int_A|u_n| <\epsilon$, for all $n$.
This is the case here: $$ \int_A |u_n|\leq \sup_n \|u_n\|_\infty |A|, $$ so $\delta=\epsilon/(M+1)$ works.
This is a particular case of a more general result, namely that if the sequence is $u_n$ is bounded in $L^1(\Omega)$, uniformly integrable, and $\Omega$ has finite Lebesgue measure, then you can extract a weakly converging subsequence $\tilde{u}_n$ in $L^1$. This is proved here in these lecture notes for example, page 37 (prop 80).
By uniqueness of the weak limit, you then get rid of the subsequence (as $u_n(t)\to u(t)$ globally in $H^{-1}$).