Suppose that $\mathcal{H}_1$ and $\mathcal{H}_2$ are two separable Hilbert spaces and that $X\subset \mathcal{H}_1$ is a dense subspace (i.e. $\overline{X}=\mathcal{H}_1$). If $\operatorname{W}:X \to \mathcal{H}_2$ is a linear surjective operator such that $$ \langle \operatorname{W}x, \operatorname{W}y\rangle_{\mathcal{H}_2} = \langle x, y\rangle_{\mathcal{H}_1} \ \forall \ x,y\in \mathcal{H}, $$ then of course $\operatorname{W}$ is a unitary operator. Now since $\overline{X}=\mathcal{H}_1$ we can extend $\operatorname{W}$ to the whole space $\mathcal{H}_1$, I do understand this last part, however the proof that I am reading also asserts that $\operatorname{W}:\mathcal{H}_1 \to \mathcal{H}_2$ is also a unitary operator and hence $\operatorname{W}^*\operatorname{W}=\operatorname{W}\operatorname{W}^*=\operatorname{I}$. I have trouble understanding that last part since we first had that $\operatorname{W}(X)=\mathcal{H}_2$, so how can we have that $\operatorname{W}^*:\mathcal{H}_2 \to \mathcal{H}_1$ is the inverse of $\operatorname{W}$?
All this comes from a proof of the spectral theorem on diagonalization of normal operators, and the argument is from Carlos S. Kubrusly: Spectral Theory of Operators on Hilbert Spaces. I am adding here a screen shot (p. 72) from the actual part proof where the operator $\operatorname{W}$ is extended, maybe there is something that I am missing. Any kind of help or insight will be very appreciated.
