Suppose U1,,UmU_1,\dots,U_m are finite dimensional subspaces of VV. Prove that U1++UmU_1 + \dots + U_m is finite dimensional and

dim(U1++Um)dimU1++dimUm\dim(U_1+\dots+U_m) \le \dim U_1 + \dots + \dim U_m

Let u1j,,udimUjju^j_1,\dots,u^j_{\dim U_j} be a basis of UjU_j for all 1jm1 \le j \le m. Concatinating all the bases gives a list of length dimU1++dimUm\dim U_1 + \dots + \dim U_m which spans U1++UmU_1 + \dots + U_m thus it's finite dimensional. Then apply 2.31 to remove dependent vectors until we have a basis for U1++UmU_1 + \dots + U_m. Thus

dim(U1++Um)dimU1++dimUm\dim(U_1+\dots+U_m) \le \dim U_1 + \dots + \dim U_m