I will copy the proof of the fact that if $A$ is in $\mathcal{B}(\mathcal{H})$ for a Hilbert space $\mathcal{H}$, then $A^{**} = A$.
For (a) we first note that by the definition of the adjoint we have $\langle A^*x,y\rangle = \langle x,A^{**}y\rangle$ for all $x$ and $y$ in $\mathcal{H}$. Since also $\langle Ay,x\rangle = \langle y,A^*x\rangle$, taking conjugates we see that $\langle x,Ay\rangle = \langle A^*x,y\rangle$. Thus $\langle x,A^{**}y\rangle = \langle x,Ay\rangle$ for all $x$ and $y$, or equivalently $\langle x,A^{**}y-Ay\rangle = 0$ for all $x$ and $y$. This forces $A^{**}y = Ay$ for all $y\in\mathcal{H}$, giving (a).
I understand more or less everything here except for the last part. I do not understand how $\langle x,A^{**}y-Ay\rangle = 0$ for all $x$ and $y$ "forces" that $A^{**}y=Ay$. Any help is appreciated.
For all $y\in\mathcal{H}$, take $x=A^{\ast\ast}y-Ay$. This says that $$ \|(A^{\ast\ast}-A)y\|=0,\forall y\in\mathcal{H}. $$ In particular, $A^{\ast\ast}-A=0$.