The 9 pseudoline counterexample means my prior definition of linear space is inadequate for constructibility. A sufficient condition for constructibility is that the space has a cospace. Then the proof of constructibility is perhaps possible as follows.
If pseudolinear space S with m boundaries has cospace C then S is constructible as hyperplanes.
Assume the conclusion for spaces with only m-1 boundaries. Construct the hyperplanes for the subspace S' of S with one boundary b removed. The subspace C' of C with vertex b removed is a cospace of S'. C' is constructible, so choose a point in the superregion. That is a consruction of b, as required.
The cospace condition is also necessary because a constructible space has a constrctible cospace.
This is still not complete, because the cospace is not unique. My technique for constructing a cospace produces a particular cospace which may be different from the subspace of the cospace obtained by removing a vertex. I guess I could patch this by requiring the space to have one of the cospaces that my cospace construction would produce, but I have to express that without reference to construction. Another option that might exist is to prove that every cospace of a constructible space is constructible.