Problem
Suppose $U_1, \dots, U_m$ are finite-dimensional subspaces $V$. Prove that $U_1 + \dots + U_m$ is finite-dimensional and $$\dim (U_1 + \dots + U_m) \leq \dim U_1 + \dots + \dim U_m$$
Solution
Theorem 1.39 implies that $U_1 + \dots + U_m$ is a subspace of $V$ and theorem 2.26 therefore ensure that $U_1 + \dots + U_m$ is finite dimensional.
To prove the second part: $$\dim U_1 + \dots + U_{m-1} + U_m \stackrel{T2.43}{=} \dim (U_1 + \dots + \dim U_{m-1})+ \dim U_m $$ $$- \dim (U_1 + \dots + U_{m-1}) \cap U_m \stackrel{\spadesuit}{\leq} \dim U_1 + \dots + \dim U_{m-1} + \dim U_m$$ Where $\spadesuit$ is due to $\dim U \geq 0$ for all subspaces $U$ - see the infobox on page 19.