How to prove the existence of non-intersecting subspaces?

Rather than look at a set of subspaces, you might look at the space $\def\R{\Bbb R}(\R^n)^{n-k}$ of $n-k$-tuples of vectors, each of which span a subspace that is a candidate for$~Q$ (some of them are linearly dependent, in which case $Q$ would not have the right dimension, but these $n-k$-tuples are among the ones that we are going to exclude). Fixing some basis of$~P$, we can complete those fixed $k$ vectors with our varying $n-k$-tuple and compute the determinant, giving a polynomial function $f:(\R^n)^{n-k}\to\R$ (in fact an $n-k$-linear alternating one) that is not identically zero (since $P$ certainly does have complementary subspaces). Now the "good" $n-k$-tuples are those for which $f$ does not vanish, and these form an open dense subset of $(\R^n)^{n-k}$. A similar construction for $P'$ gives another open dense subset, and the intersection of the two is still open and dense. In particular it is not empty, and the span of any $n-k$-tuple taken from it gives a$~Q$.

This argument immediately generalises to any finite set of $k$-dimensional subspaces to which one seeks a common complementary subspace.