Point in Hilbert Sequence Space has no Compact Neighborhood

Theorem
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Let $\ell^2 = \left({A, d_2}\right)$ be the Hilbert sequence space on $\R$.

Then no point of $\ell^2$ has a compact neighborhood.

Proof
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space.

Let $x = \left\langle{x_i}\right\rangle \in A$ be a point of $\ell^2$.

Consider the closed $\epsilon$-ball of $x$ in $\ell^2$:
 * ${B_\epsilon}^- \left({x}\right) := \left\{{y \in A: d \left({x, y}\right) \le \epsilon}\right\}$

for some $\epsilon \in \R_{>0}$.

Consider the point:
 * $y_n = \left({x_1, x_2, \ldots, x_{n - 1}, x_n + \epsilon, x_{n + 1}, \ldots}\right)$

We have that $y_n \in {B_\epsilon}^- \left({x}\right)$.

But for $m \ne n$ we have that:
 * $d \left({y_m, y_n}\right) = \epsilon \sqrt 2$

and so $\left\{ {y_n}\right\}$ has no convergent subsequence.

Thus ${B_\epsilon}^- \left({x}\right)$ is not compact.

So $x$ has no compact neighborhood.