Proving the dimension formula

Setup. Let $(u_1,\ldots,u_k)$ be a basis of $F\cap G$, with $k=\dim(F\cap G)$. Extend it to a basis $(u_1,\ldots,u_k,v_1,\ldots,v_p)$ of $F$ (so $p=\dim F-k$) and to a basis $(u_1,\ldots,u_k,w_1,\ldots,w_q)$ of $G$ (so $q=\dim G-k$).

Span. The family $(u_1,\ldots,u_k,v_1,\ldots,v_p,w_1,\ldots,w_q)$ spans $F+G$, because every element of $F$ is a combination of the $u$'s and $v$'s, and every element of $G$ is a combination of the $u$'s and $w$'s.

Independence. Suppose $\sum\alpha_i u_i+\sum\beta_j v_j+\sum\gamma_l w_l=0$. Then $\sum\gamma_l w_l=-\sum\alpha_i u_i-\sum\beta_j v_j\in F$, so it belongs to $F\cap G$, hence is a combination of the $u$'s. But the $w$'s are independent from the $u$'s, so $\gamma_l=0$ for all $l$. Then $\sum\alpha_i u_i+\sum\beta_j v_j=0$, which gives $\alpha_i=\beta_j=0$ by independence in $F$.

Conclusion. The family is a basis of $F+G$, with $k+p+q=\dim F+\dim G-\dim(F\cap G)$ vectors.