Proof that the canonical mapping of $V$ into $V^{**}$ is not bijective Let $V$ be an infinite vector space over a skew field $K$ and $V^{**}$ its bidual. Write $c_V$ for the canonical injection of $V$ into $V^{**}$. I want to show that $c_V$ is not surjective.
We can restrict to the case where $V:= K_s^{(L)}$ for some infinite set $L$. In that case there is an isomorphism of $\varphi$ of $V^*$ onto $K^L$. Let $(a_\lambda)_{\lambda\in L}$ be the canonical basis of $V$ and $(a^*_\lambda)_{\lambda\in L}$ be the corresponding family of coordinate forms in $V^*$. Let $F'$ be the span of the family $(a^*_\lambda)_{\lambda\in L}$ in $V^*$. Then $V^*\ne F'$, otherwise $K^{(L)}_d=\varphi(F')=\varphi(V^*)=K^L$. This implies that there exists a hyperplane $H'$ of $V^*$ such that $F'\subset H'$. Clearly $H'\ne V^*$; it follows that $(V^*/H')^*\ne0$ and $(V^*/H')^*\cong H''$ where $H''$ is the orthogonal of $H'$ in $V^{**}$. Write $F$ for the orthogonal of $F'$ in $V$. At this point the author says that
$$(H''\cap c_V(V))\subset c_V(F)=0.$$
I don't understand why $c_V(F)=0$? By defintion
$$F:=\{x\in V\ |\ (\forall x^*)(x^*\in F'\implies\langle x,x^*\rangle=0)\}.$$
Take $x\in F$ and $x^*\in V^{*}$. Why would $c_V(x)(x^*)=\langle x,x^*\rangle=0$?

This is just because $F$ itself is $0$.  Indeed, if $x\in F$, then in particular $\langle x,a_\lambda^*\rangle=0$ for each $\lambda$, which means every coordinate of $x$ is $0$ so $x$ is $0$.
