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 $\ell^2 = \struct {A, d_2}$ 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 $x = \sequence {x_i} \in A$ be a point of $\ell^2$.

Consider the closed $\epsilon$-ball of $x$ in $\ell^2$:
 * $\map { {B_\epsilon}^-} x := \set {y \in A: \map {d_2} {x, y} \le \epsilon}$

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

Consider the point:
 * $\sequence {y_n} = \tuple {x_1, x_2, \ldots, x_{n - 1}, x_n + \epsilon, x_{n + 1}, \ldots}$

We have that $\sequence {y_n} \in \map { {B_\epsilon}^-} x$.

But for $m \ne n$ we have that:
 * $\map {d_2} {y_m, y_n} = \epsilon \sqrt 2$

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

Thus $\map { {B_\epsilon}^-} x$ is not compact.

So $x$ has no compact neighborhood.