Suppose $U_1,\dots,U_m$ are finite dimensional subspaces of $V$ such that $U_1 + \dots + U_m$ is a direct sum. Prove that $U_1 \oplus \dots \oplus U_m$ is finite dimensional and

The short proof that I think the book intended is to use inclusion-exclusion (the general form of 2c/17) combined with the intersection of any $U_j$'s being $\{0\}$. Below is a longer proof of a stronger result.

Pick a basis for each $U_j$ then concatinate all the bases together. I claim this new list is a basis for $U_1 \oplus \dots \oplus U_m$.

Let $u^j_1,\dots,u^j_{\dim U_j}$ denote the basis of $U_j$, our list of bases is (for $d_j = \dim U_j$)

(I'm slightly abusing notation treating a list of lists the same as it's flattened coutnerpart)

To show this big list is a basis for $U_1 \oplus \dots \oplus U_m$ we need to show spanning and independence. Spanning is obvious so let's turn to independence, we want to show the only way for

To happen is for every $a^j_k = 0$. We can chunk this into a sum $u_1 + \dots + u_m$ where

Apply 1.44 to conclude $u_j = 0$, which implies $a^j_k = 0$ for all $1 \le k \le d_j$. Applying this for every $1 \le j \le m$ gives every $a^j_k = 0$ as desired. Thus the concatinated list is a basis, and as a corralary

Lots of notation to illustrate a simple idea in my mind, perhaps labeling the bases for $u$'s would be better? if I had let $B_j$ be the list of basis vectors for $U_j$ then I could have taken a combination like

Where $A_j$ is a list of coefficents for the $B_j$ basis vectors, and multiplication is defined as the usual multiply-and-sum.

I'm not sure if extra notation like this would have helped understanding, it's hard for me to evaluate how confusing notation is when I'm the one who invented it.