Problem
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 $$\dim (U_1 \oplus \dots \oplus U_m) = \dim U_1 + \dots + \dim U_m$$
Solution
Proceed by induction on $m$
BC: for $m = 1$ we have $\dim U_1 = \dim U_1$
IH: suppose that it holds for some $m = k \in \mathbb{N}$ that $\dim (U_1 \oplus \dots \oplus U_k) = \dim U_1 + \dots + \dim U_k$
IS: then it follows that: $$\dim (U_1 \oplus \dots \oplus U_k \oplus U_{k+1}) \stackrel{T2.43}{=} \dim (U_1 \oplus \dots \oplus U_k) + \dim U_{k+1}$$ $$- \dim(U_1 \oplus \dots \oplus U_k) \cap U_{k+1} \stackrel{\spadesuit}{=} \dim (U_1 \oplus \dots \oplus U_k) + \dim U_{k+1}$$ $$\stackrel{IH}{=} \dim U_1 + \dots + \dim U_k + \dim U_{k+1}$$ Where $\spadesuit$ is due to $U_1 \oplus \dots \oplus U_k \oplus U_{k+1}$ being a direct sum.