Courant–Fischer Theorem Proof [Meyer]. Intersection of two subspaces with same dimension. Suppose we have 2 subspaces $\mathcal{A} \subseteq \mathbf{R}^n$ and $\mathcal{B}\subseteq \mathbf{R}^n$ that have the same dimension, say dim $\mathcal{A}=$ dim $\mathcal{B}=l>0$. Is it true that they have a non null intersection?

Ignore this part and focus on the top question answered correctly by @paf

I'm reading a section of Meyer's book, Matrix Analysis and Applied Linear Algebra, on the Courant–Fischer Theorem. I'm following the proof of pages 550-551, but I don't seem to understand why $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$ . . . This two subspaces have the same dimension so I figure it might be a general property that it's being used. Is this the case? If not, how would your show that $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$, following the reasoning of Meyer? 

I'm not sure of your question, hence I'll give two answers.
1. Two vector subspaces of $\Bbb R^n$ always have non-empty ($\ne \varnothing$) intersection. Indeed, by definition, every vector subspace must contain the zero vector.
2. But their intersection can be exactly $\{0\}$ (thus having dimension $0$). Simply take two distinct vector subspaces of dimension 1 in $\Bbb R^2$ (i.e. 2 lines through 0)! To have an example valid for all $n$, if $V = \{a_1x_1 + \dots + a_nx_n = 0\}\subset \Bbb R^n$, then the line $W$ generated by $(a_1,\dots,a_n)$ intersects $V$ only at the zero vector.
So don't confuse $\varnothing$ and $\{0\}$...
