We have $2^*=\frac{2N}{N-2}>\frac{2(N-2)}{N-2}=2\geq 1$ so $u_{n_k}\to u_0$ in $L^2$. Since the sequence $\{u_{n_k}\}$ is bounded in $L^2$, we can extract a converging subsequence $\{u_{\psi(k)}\}$ in $L^2$ to a function $v$ (taking again a subsequence we can assume it converges almost everywh. Using test functions and weak convergence, we can see that in fact $v=u$.
We show the following result
If $\{f_n\}\subset H^1_0(\Omega)$ is a sequence which converges weakly to $f$ in $H^1_0(\Omega)$, then this sequence converges weakly to $f$ in $L^2(\Omega)$.
Since $\{ f_n\}$ and $\{ \nabla f_n\}$ are bounded in $L^2$, we can extract converging subsequences $\{f_{\psi(n)}\}$ and $\{\nabla f_{\psi(n)}\}$, which converges respectively to $g$ and $h$. But for $\varphi\in\mathcal D(\Omega)$ and $1\leq i\leq N$ $\int_{\Omega}gD_i\varphi dx=\lim_k\int_{\Omega}f_{\psi(k)}D_i\varphi dx =-\lim_k\int_{\Omega}D_if_{\psi(k)}\varphi dx =-\int_{\Omega}h_i\varphi dx, $ so $h=\nabla g$ and $f_{\psi(k)}$ converges to $g$ weakly in $H^1_0(\Omega)$, so $f=g$.
So the sequence $\{u_n\}$ admit a subsequence which converges weakly to $u$ and $u_0$, which implies that $\langle w,u-u_0\rangle=0$ for each $w\in H$, so $u=u_0$.