What I did:
Using the theorem of sum of subspaces I have $dim(U \oplus W) = dim(U) + dim(W)$
It's not hard to verify that $U \cap W = \{ 0 \} \Rightarrow \{u_{1},...,u_{n},w_{1},...,w_{m}\}$ is l.i. ($n=dim(U),\ m=dim(W)$)
Assume $dim(V) < dim(U)+dim(W)$
As 2 l.i. subspaces have together greater dimension than the $V$ space itself, there exists a vector of $[u_{1},...,u_{n},w_{1},...,w_{m}]$ that can't be written as a linear combination of $V$ basis vectors (verifiable). Which leads to a contradiction.
I think maybe there's a more simple solution, if anyone knows post it here please.