Infinite orthonormal set and space with inner product that is not compact

Question

Let $X$ be a space with inner product and $S$ be an infinite orthonormal set in $X$. Prove that $S$ is not compact but is closed and limited.

Answer 1

If $\{e_1,e_2,...\}$ is a sequence of distinct points in $S$ then $\|e_n-e_m\|=\sqrt2 $ for $n \neq m$ so the sequence contains no convergent subsequence. Hence $S$ is not compact. It is obvious that $S$ is bounded. If a sequence from $S$ converges to $x$ then it ,must be one member of the sequence because of the property $\|e-f\|=\sqrt2 $ for $e \neq f$ in $S$. Hence $S$ is closed.

[Suppose $x_n \in S$ for all $n$ and $x_n \to x$. Then $\|x_n-x_m| <\sqrt2 $ for $n$ and $m$ sufficiently large which implies that $x_n=x_m$ for $n$ and $m$ sufficiently large. Hence $x=x_n \in S$ if $n$ is sufficiently large].