To obtain a contradiction, suppose that $b(u_n)\not\to b(u)$. Passing to a subsequence, without loss of generality (wlog) we have $\|b(u_n)-b(u)\|\ge c$ for some real $c>0$ and all $n$.

By the mean value theorem, $b(u_n)-b(u)=b'(v_n)(u_n-u)$ for all $n$ and some $v_n$ on the straight line segment from $u$ to $u_n$.

Since the sequence $(u_n)$ is weakly convergent, it is bounded. So, recalling that $b'$ is compact and passing to a subsequence, wlog we have $b'(v_n)\to B$ for some $B\in E^*$. Also, $B(u_n-u)\to0$, since $u_n\to u$ weakly. Also, $\|b'(v_n)-B\|\,\|u_n-u\|\to0$, since $b'(v_n)\to B$ and the sequence $(u_n)$ is bounded. So, $$|b(u_n)-b(u)|=|b'(v_n)(u_n-u)|\le\|b'(v_n)-B\|\,\|u_n-u\|+|B(u_n-u)|\to0+0=0,$$ which contradicts the condition that $\|b(u_n)-b(u)\|\ge c$ for some real $c>0$ and all $n$. $\quad\Box$

To obtain a contradiction, suppose that $b(u_n)\not\to b(u)$. Passing to a subsequence, without loss of generality (wlog) we have $|b(u_n)-b(u)|\ge c$ for some real $c>0$ and all $n$.

By the mean value theorem, $b(u_n)-b(u)=b'(v_n)(u_n-u)$ for all $n$ and some $v_n$ on the straight line segment from $u$ to $u_n$.

Since the sequence $(u_n)$ is weakly convergent, it is bounded. So, recalling that $b'$ is compact and passing to a subsequence, wlog we have $b'(v_n)\to B$ for some $B\in E^*$. Also, $B(u_n-u)\to0$, since $u_n\to u$ weakly. Also, $\|b'(v_n)-B\|\,\|u_n-u\|\to0$, since $b'(v_n)\to B$ and the sequence $(u_n)$ is bounded. So, $$|b(u_n)-b(u)|=|b'(v_n)(u_n-u)|\le\|b'(v_n)-B\|\,\|u_n-u\|+|B(u_n-u)|\to0+0=0,$$ which contradicts the condition that $|b(u_n)-b(u)|\ge c$ for some real $c>0$ and all $n$. $\quad\Box$