Let $A \in M_{p\times q}(\mathbb K)$, $C \in M_{(n-p)\times q}(\mathbb K)$ and $B \in M_{(n-p)\times (n-q)}(\mathbb K)$. We may decompose $\mathbb K^n=\mathbb K^p\times 0 \oplus 0 \times \mathbb K^(n-p)$ and consider $\{a_1,\dots,a_q\} \subset \mathbb K^p\times 0$ the set of columms of $A$, $\{b_1,\dots,b_q\} \subset 0 \times \mathbb K^{n-p}$ the set of columms of $B$ and $\{c_1,\dots,c_q\} \subset 0 \times \mathbb K^{n-p}$ the set of columms of $C$.
So, it suffices to show that given an LI subset $\{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ of $\{a_1,\dots,a_q,b_1,\dots,b_{n-q}\}$ we have that $\{a_{i_1}+c_{i_1},\dots,a_{i_s}+c_{i_s},b_{j_1},\dots,b_{j_r}\}$ is a LI subset of $\{a_1+c_1,\dots,a_q+c_q,b_1,\dots,b_{n-q}\}.$ In fact, if $\alpha_{i_1}(a_{i_1}+c_{i_1})+\dots+\alpha_{i_s}(a_{i_s}+c_{i_s})+\beta_{j_1}b_{j_1},\dots,\beta_{j_r}b_{j_r} = 0$ for some $\alpha$s and $\beta$s \in $\mathbb K$, we must have that $\mathbb K^p \times 0 \ni \alpha_{i_1}a_{i_1} +\dots+\alpha_{i_s}a_{i_s} = 0$ and $0 \times \mathbb K^{n-p} \ni \alpha_{i_1}c_{i_1}+\dots+\alpha_{i_s}c_{i_s}+\beta_{j_1}b_{j_1},\dots,\beta_{j_r}b_{j_r} = 0.$ By the linear independence of $\{a_{i_1},\dots,a_{i_s}\} \subset \{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ we must have that $\alpha_{i_1}=\dots=\alpha_{i_s}=0$ and, hence, $\beta_{j_1}b_{j_1}+\dots+\beta_{j_r}b_{j_r} = 0.$ Again, $\{b_{j_1},\dots,b_{j_r}\}\subset \{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ implies that $\beta_{j_1}=\dots=\beta_{j_r}=0.$ Therefore, we conclude that $\{a_{i_1}+c_{i_1},\dots,a_{i_s}+c_{i_s},b_{j_1},\dots,b_{j_r}\}$ is a LI set.