Let the columns of $A$ and $B$ be $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ respectively. By definition, the rank of $A$ and $B$ are the dimensions of the linear spans $\langle a_1, \ldots, a_n\rangle$ and $\langle b_1, \ldots, b_n\rangle$. Now the rank of $A + B$ is the dimension of the linear span of the columns of $A + B$, i.e. the dimension of the linear span $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$. Since $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$ the result follows.
Edit: Let me elaborate on the last statement. Any vector $v$ in $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$ can be written as some linear combination $v = \lambda_1 (a_1 + b_1) + \ldots + \lambda_n (a_n + b_n)$ for some scalars $\lambda_i$. But then we can also write $v = \lambda_1 (a_1) + \ldots + \lambda_n (a_n) + \lambda_1 (b_1) + \ldots + \lambda_n (b_n)$. This implies that also $v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$. We can do this for any vector $v$, so
$$\forall v \in \langle a_1 + b_1, \ldots, a_n + b_n\rangle: v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$$
This is equivalent to saying $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$.
Best Answer
Hint: Let $e_1,\ldots,e_n$ be the standard basis of $\Bbb R^n$, then the spanned subspaces satisfy: $$\langle \ldots, (A+B)e_i,\ldots \rangle \subseteq \langle \ldots, Ae_i,\ldots \rangle +\langle \ldots,Be_i,\ldots\rangle .$$