Finding a subspace whose intersections with other subpaces are trivial. On p.24 of the John M. Lee's Introduction to Smooth Manifolds (2nd ed.), he constructs the smooth structure of the Grassmannian. And when he tries to show Hausdorff condition, he says that for any 2 $k$-dimensional subspaces $P_1$, $P_2$ of $\mathbb{R}^n$, it is always possible to find a $(n-k)$-dimensional subspace whose intersection with both $P_1$ and $P_2$ are trivial.
My question: I think it is also intuitively obvious that we can always find $(n-k)$-dimensional subspace whose intersection with m subspaces  $P_1,\ldots,P_m$ are trivial, which is a more generalized situation. But I can't prove it rigorously. Could you help me for this generalized case?
Thank you.

For $j=1,2,\ldots,m$, let $B_j$ be a basis of $P_j$. It suffices to find a set of vectors $S=\{v_1,v_2,\ldots,v_{n-k}\}$ such that $S\cup B_j$ is a linearly independent set of vectors for each $j$. We will begin with $S=\phi$ and put vectors into $S$ one by one.
Suppose $S$ already contains $i<n-k$ vectors. Now consider a vector $v_{i+1}=v_{i+1}(x)$ of the form $(1,x,x^2,\ldots,x^{n-1})^T$. For each $j$, there are at most $n-1$ different values of $x$ such that $v_{i+1}(x)\in\operatorname{span}\left(\{v_1,v_2,\ldots,v_i\}\cup B_j\right)$, otherwise there would exist a non-invertible Vandermonde matrix that corresponds to distinct interpolation nodes. Therefore, there exist some $x$ such that $v_{i+1}(x)\notin\operatorname{span}\left(\{v_1,v_2,\ldots,v_i\}\cup B_j\right)$ for all $j$. Put this $v_{i+1}$ into $S$, $S\cup B_j$ is a linearly independent set. Continue in this manner until $S$ contains $n-k$ vectors. The resulting $\operatorname{span}(S)$ is the subspace we desire.
