Let $(\mathcal{H},\langle{,}\rangle)$ be a separable, infinite-dimensional Hilbert space. Let $\mathcal{X}''$ denote the space of bounded sequences in $\mathcal{H}$. For a Banach limit $L$, define a bilinear function $\mathcal{X}''\times\mathcal{X}''\rightarrow\mathbb{C}$ by
$$\langle{\left\{x_{n}\right\},\left\{y_{n}\right\}}\rangle:=L \langle{x_{n},y_{n}}\rangle$$
Let $\mathcal{X}'$ be the inner product space obtained by modding out by the equivalence relation $\left\{x_{n}\right\}\sim\left\{y_{n}\right\}\Leftrightarrow L\left\|x_{n}-y_{n}\right\|^{2}=0$. Using the separability of $\mathcal{H}$, it is not hard to show that $\mathcal{X}'$ has cardinality of the continuum $\mathfrak{c}$ and $\mathcal{X}'$ contains an orthonormal system $\left\{\psi_{\alpha}\right\}$ of cardinality $\mathfrak{c}$; so $\mathcal{X}'$ is inseparable.
In the paper "Two-Sided Ideals and Congruences in the Ring of Bounded Operators in Hilbert Space", the author gives the following argument for the incompleteness of $\mathcal{X}'$: For an orthonormal system $\left\{\psi_{\alpha}\right\}$ with cardinality $\mathfrak{c}$, the space $\mathcal{X}$ of all series
$$\sum_{\alpha}a_{\alpha}\psi_{\alpha}\text{ with }\sum_{\alpha}\left|a_{\alpha}\right|^{2}<\infty$$ has cardinality $2^{\mathfrak{c}}$ and therefore is not a subset of $\mathcal{X}'$, which has cardinality $\mathfrak{c}$.
I don't follow this argument. A series $\sum_{\alpha\in A}a_{\alpha}\psi_{\alpha}$ converges only if at most countably many coefficients are nonzero. So wouldn't there be a surjection $A^{\mathbb{N}}\times\ell^{2}(\mathbb{N})\twoheadrightarrow\mathcal{X}$ and therefore
$$\text{card }\mathcal{X}\leq\max\left\{\text{card }A^{\mathbb{N}}, \text{card }\ell^{2}(\mathbb{N})\right\}=\max\left\{2^{\aleph_{0}\times\aleph_{0}},\mathfrak{c}\right\}=\mathfrak{c}$$
Given that the author attributes the aforementioned proof to von Neumann, I feel I must have made a mistake. Would someone please help me out?
You are right. For an infinite set $S$, we have
$$\ell^2(S) = \bigcup_{\substack{T\subset S\\\operatorname{card} T = \aleph_0}} \ell^2(T),$$
viewing $\ell^2(T)$ as a subset of $\ell^2(S)$ via the canonical embedding, since a summable family of real (or complex) numbers can only contain countably many nonzero terms. Since $\operatorname{card} \ell^2(T) = \mathfrak{c}$ for all countably infinite $T$, we have
$$\operatorname{card} \ell^2(S) \leqslant \mathfrak{c} \cdot \operatorname{card} \{ T\subset S : \operatorname{card} T = \aleph_0\}.$$
But every countably infinite subset $T$ of $S$ is the image of a map $t\colon \mathbb{N}\to S$, so
$$\operatorname{card} \{ T\subset S : \operatorname{card} T = \aleph_0\} \leqslant \operatorname{card} \bigl(S^\mathbb{N}\bigr) = (\operatorname{card} S)^{\aleph_0}.$$
With $\operatorname{card} S \leqslant \mathfrak{c}$, we thus get
$$\operatorname{card} \ell^2(S) \leqslant \mathfrak{c}\cdot \mathfrak{c}^{\aleph_0} = \mathfrak{c}.$$