A sequence with finite inner product with any element of $\ell^2$ is itself in $\ell^2$

Question

$\ell^2$ here is real, not complex. Consider a sequence $x:\mathbb{N}\to\mathbb{R}$. For any element $y\in\ell^2$, we have $(x,y)_{\ell^2}<\infty$. How do we prove that $x\in\ell^2$? We can assume all the entries of $x$ and $y$ above are nonnegative. If we can show $(x,\cdot)$ is bounded on $\ell^2$, the statement follows from Riesz rep. theorem, though.

Answer 1

Let $B_k = \{y \in \ell^2: |(x,y)| \le k\}$. These are closed, and their union is $\ell^2$. By the Baire Category Theorem some $B_k$ has nonempty interior. This implies that $(x, \cdot)$ is bounded.