Given $3$-dimensional subspaces $V, W \subset \Bbb R^5$, there is a nonzero vector in $V \cap W$

Let {v1,v2,v3} - any basis of V;

Let {w1,w2,w3} - any basis of W;

vectors {v1,v2,v3,w1,w2,w3} are linearly dependent, because 6 vectors are too much for R5.

So that, there exist linear combination

c1*v1+c2*v2+c3*v3+d1*w1+d2*w2+d3*w3=0

such that coefficients c1,c2,c3,d1,d2,d3 are NOT ALL zeros.

Thus, common vector:

t = c1*v1+c2*v2+c3*v3 = -d1*w1-d2*w2-d3*w3

in {v1,v2,v3} and {w1,w2,w3} basis respectively.

t is zero-vector iff c1,c2,c3,d1,d2,d3 are ALL zeros.

But prevoiusly showed that they are NOT ALL zero, so t is not zero-vector.

HINT: $$ \dim (V + W) = \dim (V) + \dim (W) - \dim (V \cap W) $$

We know that if $V$ and $W$ are subspaces, then $V\cap W$ and $V+W$ are also subspaces. Recall $V+W$ is the smallest subspace which contains both $V$ and $W$. So, $\dim(V+W)$ is 3 or 4 or 5.

We know that $\dim(\mathbb{R}^5) \geq \dim(V+ W) = \dim(V)+\dim(W)-\dim(V\cap W)$ which implies that $5 \geq \dim(V+ W) = 3+3-\dim(V\cap W)$.

Hence, to satisfy above equation, $\dim(V\cap W)$ is at least 1. Thus, any basis of $V\cap W$ contains at least one non-zero vector, i.e., $V\cap W=\text{span}(S)$, where $|S|\geq 1$, $S$ being basis. Take any vector in $S$ or non-zero vector in $\text{span}(S)$ vector.