Step 1: Use a complement-counting argument instead of the addition formula directly. Extend a basis of $U \cap W$ to a basis of $U$, and separately to a basis of $W$; let $k = \dim(U\cap W)$. Then $U$ contributes $4-k$ basis vectors beyond $U\cap W$, and $W$ contributes $5-k$ basis vectors beyond $U\cap W$. Together with the $k$ shared vectors, $U+W$ is spanned by $(4-k)+(5-k)+k = 9-k$ linearly independent vectors, so $\dim(U+W) = 9-k$.
Step 2: Since $U+W \subseteq V$ and $\dim V = 7$, we need $9-k \le 7$, giving $k \ge 2$.
Step 3: Also $U \cap W \subseteq U$ forces $k \le \dim U = 4$.
Step 4: So $k \in \{2,3,4\}$ are the only values consistent with these constraints, and each is realizable. For instance in $\mathbb{R}^7$ with basis vectors $e_1,\dots,e_7$, let $U = \text{span}(e_1,e_2,e_3,e_4)$. Taking $W=\text{span}(e_1,e_2,e_3,e_4,e_5)$ gives $U\cap W=U$, dimension 4. Taking $W=\text{span}(e_2,e_3,e_4,e_5,e_6)$ gives $U\cap W=\text{span}(e_2,e_3,e_4)$, dimension 3. Taking $W=\text{span}(e_3,e_4,e_5,e_6,e_7)$ gives $U\cap W=\text{span}(e_3,e_4)$, dimension 2.
Step 5: No construction can push $k$ down to 1, since that would force $\dim(U+W)=8$, exceeding $\dim V = 7$, which is impossible. \[\boxed{1}\]