Why does the union of $k$-dimensional subspaces contain no „new“ $k$-dimensional subspaces?

Question

For simplification, let $V$ be a finite-dimensional, real vector space. I know that I cannot represent $V$ as the union of finitely many proper subspaces (or even just $k-1$-dimensional subspaces). But how about the following:

Let $W_1,…,W_n\leq V$ be $k$-dimensional subspaces. Then every $k$-dimensional subspace of $W_1\cup \ldots \cup W_n$ must coincide with one of the $W_i$.

My initial idea was to choose a basis and trace where in which $W_i$ they could possibly „land“ on the right hand side, but I didn't find a proof based on this.

Answer 1

Two non-equal $k$-dimensional subspaces must intersect in a $k-1$-dimensional subspace. If some new subspace were to exist, it would have to be the union of $k$ $k-1$ dimensional subspaces, which as you have correctly stated is impossible.