 Solution

Choose a basis $e_1,\dots ,e_k$ for $U\cap U^{\prime }$. Extend it to a basis $e_1,\dots ,e_k,u_1,\dots ,u_r$ of $U$, and separately to a basis $e_1,\dots ,e_k,v_1,\dots ,v_s$ of $U^{\prime }$. The combined list spans $U+U^{\prime }$.

To prove linear independence, suppose a linear combination of the combined list is zero. Then the combination of the $v_j$ lies in $U\cap U^{\prime }$, so it can be expressed using the $e_i$. Independence of the chosen basis of $U^{\prime }$ forces every $v_j$ coefficient to vanish. Independence of the chosen basis of $U$ then forces the remaining coefficients to vanish. Thus the combined list is a basis, with $k+r+s$ elements. This proves the dimension formula for a sum of subspaces: