Convergence in $C([0,T_0],L^2)$ and uniform boundedness in $C([0,T_0],H^2)$ gives convergence in $C([0,T_0],H^1)$.

I am not exactly sure how to apply Aubin-Lions here, as suggest by BibgearZzz. However, I think what the authors mean by ‘interpolating the given convergence with the uniform bound estimates’ is the following.

First note that $(v_n(t))_{n\in\mathbb{N}}$ has a subsequence that converges to $v(t)$ weakly in $H^{s+1}(\Omega)$. In particular, $v(t)\in H^{s+1}(\Omega)$ and $\|v(t)\|_{H^{s+1}}\leq M$. Then you can use that for every $\epsilon>0$ there exists $C>0$ such that $$ \|f\|_{H^{s}}\leq \epsilon\|f\|_{H^{s+1}}+C\|f\|_{L^2} $$ for all $f\in H^{s+1}$ (I don't know a name for this inequality, but it's proof boils down to partial integration + Young's inequality - it's Theorem 7.28 in Gilbarg-Trudinger).

Applied to the case at hand, you get $$ \|v_n(t)-v(t)\|_{H^s}\leq \epsilon\|v_n(t)-v(t)\|_{H^{s+1}}+C\|v_n(t)-v(t)\|_{L^2}\leq 2\epsilon M+C\|v_n-v\|_{C([0,T_0];L^2)}. $$ Letting $n\to\infty$ and then $\epsilon\searrow 0$ yields the desired conclusion.