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.