Let $E$ be a vector space, and $F_i$, $1\leq i \leq s$ subspaces of $E$ of dimension $n_i$ and $b_i = (e_{i,1}, \dots, e_{i,n_i})$ a base of $F_i$ and $F = F_1 \oplus \dots \oplus F_n$. Show that the concatenation of $b_i$ forms a base of $F$ and that $\dim F = \sum n_i = \sum \dim F_i$.
My attempt:
Let $u \in F$. Then $u = f_1 + \dots +f_s$ with $f_i = \lambda_{i,1}e_{i,1} + \dots + \lambda_{i,n_i}e_{i,n_i}$.
Thus $$u = \lambda_{1,1}e_{1,1} + \dots + \lambda_{1,n_i}e_{1,n_i} + \dots \dots \dots + \lambda_{s,1}e_{s,1} + \dots + \lambda_{s,n_i}e_{s,n_i}$$
$u$ is written in a unique way using $f_i$ who themselves are written in a unique way using the elements of the basis $b_i$. Thus the concatenation of the $b_i$ forms a basis of $F$. And so we have $\dim F = {\rm card}\,(b) = \sum \dim(b_i)$.
Is my approach correct? I am rather really unsure.