If $U$ and $W$ are subspaces of $V$ whose dimension is $9$, and $\dim(U) = 3$, and $\dim(W) = 5$, what could be the possible values of $\dim(U \cap W)$?
By thinking about it it seems the possible values are $0, 1, 2, 3$ because the intersection could not possible be more than the dimension of the smallest one, right?
If my answer is correct, how do I formally prove this?
By dim(U \ W) I assume you mean $\dim(U \cap W)$, the dimension of the intersection? If that's the case then yes, you are correct, the possible dimensions are $0, 1, 2, 3$.
To prove that you can't have a dimension larger than $3$ you do exactly as you have suggested, you observe that $U \cap V$ is a subspace of $U$ so $\dim(U \cap V) \leq \dim(U)$. Now to prove that $0, 1, 2, 3$ are actually possible you just give examples where this happens. Let $V = \mathbb R^9$ and $W$ the span of the first $5$ standard unit vectors. Then for each of $0, 1, 2, 3$ there is a choice of $U \subseteq \mathbb R^9$ such that the intersection has the correct dimension. I'll leave it to you to figure out what $U$ should be in those cases.
We know that $U\cap W$ is a subspace of $U$, so has dimension at most $3$. Each of the dimensions between $0$ and $3$ occurs, however. Indeed, pick a basis $\{v_1,\ldots,v_9\}$ of $V$. Let $U=span\{v_1,v_2,v_3\}$. To get $\dim(U\cap W)=j$, where $j\in\{0,1,2,3\}$, set $W:=span\{v_{4-j},\ldots,v_{9-j}\}$.
