By the continuity of the inner product, a subspace of a Hilbert space is closed? Let $(e_n)$ be an orthonormal sequence in a Hilbert space $H$. The following are equivalent:


*

*$(e_n)$ is complete.

*$\text{clin}\{e_n \ | \ n\in \mathbb{N}\} = H$,

*$||x||^2 = \sum_{n=1}^\infty |(x,e_n)|^2, \quad \text{for all} \ x\in H$.


For showing $(2)=>(1)$, the following text is used in the proof

Let $x\in H$ such that $(x,e_n)=0$ for all $n$. Define $$ E = \{y\in H
: (x,y) = 0\}. $$ $E$ is a subspace of $H$ and by the continuity of
  the inner product it is closed.

I don't understand how the continuity of the inner product is used to state $E$ is closed?

Let $(y_n)$ be a convergent sequence in $E$ and denote by $y_0$ its limit.
We have $(x,y_n)=0$ for all $n$ and, by continuity of the inner product:
$(x,y_n) \to (x,y_0)$.
This shows $ (x,y_0) =0$, hence $y_0 \in E$.
