Prove that if $U$ and $W$ are 3-dimensional subspaces of $\mathbb{R}^5$ then $U ∩ W$ is non-trivial I said that since a 3-dimensional subspace is the span of 3 linearly independent vectors then $U$ and $W$ must be linearly independent.

Since $U$ and $W$ are linearly independent then they span the entire 3-dimensional space.

Since $U$ and $W$ span the entire space, they intersect everywhere.

Therefore, $U \cap W$ is non-trivial for every vector in $\mathbb{R}^3$ except for the null vector.

This was my reasoning but can anybody tell me if my thinking is correct. I feel like I am missing something since I did not make use of $\mathbb{R}^5$.
Choose bases $A=\{u_1,u_2,u_3\}$ and $B=\{w_1,w_2,w_3\}$ for $U$ and $W$ respectively. If $A \cap B$ is nonempty we are done. Suppose not. Then $A \cup B$ contains $6$ distinct vectors in $\mathbb{R}^5$, a $5$ dimensional space, so they must be linearly dependent. This means there exist scalars $\lambda_i, \mu_i$ not all equal to zero such that
$$\sum_i \lambda_i u_i +\sum_i \mu_i w_i=0.$$
Taking the right sum to the other side we see that $\sum_i \lambda_i u_i \in W$. Since it is also in $U$, it is in $U \cap W$. It is nonzero because if it would be zero then the $\lambda_i$ would all be zero and by linear independence of the $u_i$ and the $\mu_i$ would be zero as well by linear independence of the $w_i$.  This shows $U \cap W$ contains a nonzero element.
Recall that $\dim(U\cap V) + \dim(U+V) = \dim U + \dim V=6$. If the intersection was trivial, $U+V$ would have dimension $6$, which of course is impossible.