# Linear isometry between Hilbert spaces has a closed range

Let $U$ be a linear isometry between Hilbert spaces. Why does the fact that the range of $U$ is dense imply that the range of $U$ is closed?

I am trying to understand the proof of theorem 5.4 in Conway's A Course in Functional Analysis.

In fact, the range of a linear isometry $U \colon H \rightarrow H'$ between Hilbert spaces must always be closed. If the range of $U$ is also dense then $U(H) = H'$ so $U$ is one-to-one and onto.
The reason is that the range $U(H)$ of $U$ is complete and a complete subspace of a normed space must be closed. To see that $U(H)$ is complete, let $Ux_n$ be a Cauchy sequence in $U(H)$ so $\| Ux_n - Ux_m \|_{H'} \rightarrow 0$. Since $U$ is an isometry, $\| x_n - x_m \|_{H} = \| Ux_n - Ux_m \|_{H'}$ and so $(x_n)$ is Cauchy in $H$. Since $H$ is complete, $x_n \rightarrow x$ for some $x \in H$ but then $Ux_n \rightarrow Ux$ .