Let the separable Hilbert cube $Q=\prod_{i=1}^{+\infty}[0,1]$ embed into the real Hilbert space $H=l^2(\mathbb{Z}^+)$, whose coordinate unit vectors are $\{ e_i \}_{i=1}^{+\infty}$, as the subset $\left\{ (x_1,x_2,\dots) \middle| 0 \le x_i \le \dfrac{1}{i^2}\right\}$, and its pseudo-boundary $B(Q)$ embed as $\left\{ (x_1,x_2,\dots) \middle| \exists i \in \mathbb{N}, x_i=0 \vee x_i=\dfrac{1}{i^2}\right\}$. I wonder whether $B(Q)$ is a contractible and locally contractible space.
Since $Q \backslash B(Q)$ is not open in $Q$, $B(Q)$ is not closed in $Q$. Thus $B(Q)$ can not be compact and is not homeomorphic to $Q$, unlike the case between the Hilbert unit ball and unit sphere. There do exist simple closed curves (and other higher dimensional analogues) in $B(Q)$ that are not contained in any finite dimensional subspace of $H$, e.g. the piecewise linear curve $$0 \rightarrow e_1 \rightarrow e_1+\dfrac{1}{2^2}e_2 \rightarrow \cdots \rightarrow \sum_{i=1}^{+\infty}\dfrac{1}{i^2}e_i \rightarrow \sum_{i=2}^{+\infty}\dfrac{1}{i^2}e_i \rightarrow \sum_{i=3}^{+\infty}\dfrac{1}{i^2}e_i \rightarrow \cdots \rightarrow 0$$ which however seems to be null-homotopic since it encloses a "finite area" in $B(Q)$.
