If $b\in C^1(E, \mathbb{R})$ and $b'$ is compact, then $b$ is weakly continuous — a reference request

Question

While reading the well known book Minimax Methods in Critical Point Theory with Applications to Differential Equations by Paul Rabinowitz, in the proof of a generalisation of the Mountain Pass Theorem (Theorem 5.29 in the book), I encountered the following abstract result:

Let $E$ be a real Hilbert space and consider $b\in C^1(E, \mathbb{R}$) such that $b'$ is compact. Then, $b$ is weakly continuous, i.e. if $(u_n)_n\subseteq E$ converges weakly to $u\in E$, then $b(u_n)\to b(u)$ as $n\to \infty$.

The reference for this result given in the book is

M. A. Krasnoselski, Topological methods in the theory of nonlinear integral equations, Macmillan, New York, 1964.

However, even if this is a well renowned book in the field of nonlinear analysis, I do not have access to it. Does anyone know a modern reference for this result? The proof looks nontrivial to me, at least I do not know how to approach it.

EDIT: The fact that $b'$ is compact means that if $A\subseteq E$ is bounded, then the closure of $b'(A)$ is compact. Of course, $b'$ denotes the mapping $x\mapsto b'(x)$, i.e. $b'$ associates with each $x\in E$ the Frechet derivative of $b$ at $x$, which we denote by $b'(x)\in E^{*}$.

Answer 1

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 $n$. $\quad\Box$