Let $H$ be a separable $\mathbb R$-Hilbert space and $C^2_b(H)$ denote the space of all $f\in C^2(H)$ such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are bounded. We know that $C^2_b(H)$ equipped with $$\left\|f\right\|_{C^2_b(H)}:=\left\|f\right\|_\infty+\left\|{\rm D}f\right\|_\infty+\left\|{\rm D}^2f\right\|_\infty$$ is a $\mathbb R$-Banach space.
Now consider the subset $U^2_b(H)$ of all $f\in C^2_b(H)$ such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are uniformly continuous on bounded subsets of $H$. Can we show that $U^2_b(H)$ is a closed subset of $C^2_b(H)$?
If I'm not missing, the claim is trivial whenever $H$ has the nearest-point property (e.g. if $H=\mathbb R^d$, $d\in\mathbb N$, or, more generally, if $H$ has the Heine-Borel property), since then every continuous function on $H$ into any other metric space is uniformly continuous on every bounded subset of $H$.
I think all you need here is the standard argument that a uniform limit of uniformly continuous functions is uniformly continuous.
Suppose that $f=\lim_nf_n$, with each $f_n$ uniformly continuous on bounded subsets. Let $\varepsilon>0$ and $Y\subset H$ be bounded. Then there exists $n$ with $\|f-f_n\|_{C_b^2(H)}<\varepsilon/3$. As $f_n$, $Df_n$, $D^2f_n$ are uniformly continuous on $Y$, there exists $\delta>0$ such that $\|D^kf_n(x)-D^kf_n(y)\|<\varepsilon/3$, $k=0,1,2$, whenever $x,y\\in Y$ and $\|x-y\|<\delta$.
Then, if $x,y\in Y$ with $\|x-y\|<\delta$, for $k=0,1,2$ we have \begin{align} \|D^kf(x)-D^kf(y)\| &\leq\|D^kf(x)-D^kf_n(x)\|+\|D^kf_n(x)-D^kf_n(y)\|+\|D^kf_n(y)-D^kf(y)\|\\[0.3cm] &\leq \|f-f_n\|_{C_b^2(H)}+\frac\varepsilon3+\|f_n-f\|_{C_b^2(H)}\\[0.3cm] &\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. \end{align} So $f$, $Df$, $D^2f$ are uniformly continuous.