How to prove that a Hilbert space is the directed colimit of its finite-dimensional subspaces?
Does this imply that the category of Hilbert spaces (and bounded linear maps) is the Ind-completion (see here) of that of finite-dimensional Hilbert spaces?
First things first: we need to be crystal-clear about what categories we are taking colimits in. Let $\textbf{Hilb}$ be the category of complex Hilbert spaces and bounded linear maps and let $\textbf{Vect}$ be the category of complex vector spaces and linear maps. (If you prefer, we can replace "complex" with "real" – it makes no difference to the argument.)
There is an evident forgetful functor $\textbf{Hilb} \to \textbf{Vect}$. It is faithful, of course, but more is true: given Hilbert spaces $V$ and $W$, the map $$\textbf{Hilb} (V, W) \to \textbf{Vect} (V, W)$$ is bijective if $V$ is finite-dimensional. Hence, for any diagram $V : \mathcal{J} \to \textbf{Hilb}$ where each $V (j)$ is finite-dimensional, we have natural bijections $$\textbf{Hilb} (\varinjlim V, W) \cong \varprojlim \textbf{Hilb} (V, W) \cong \varprojlim \textbf{Vect} (V, W) \cong \textbf{Vect} (\varinjlim V, W)$$ provided $\varinjlim V$ exists in both $\textbf{Hilb}$ and $\textbf{Vect}$. The naturality of this bijection means that, if $\varinjlim V$ exists in $\textbf{Hilb}$ and is preserved by the forgetful functor, then $\varinjlim V$ has the property that every linear map $\varinjlim V \to W$ is bounded. But this is impossible if $\varinjlim V$ is infinite-dimensional. So either $\varinjlim V$ does not exist in $\textbf{Hilb}$ or is not preserved by the forgetful functor.
In particular, an infinite-dimensional Hilbert space is not the colimit of its finite dimensional subspaces: the underlying vector space is, so if the Hilbert space were also, then the colimit would be preserved by the forgetful functor.