Consider the matrix $$ A\equiv \begin{pmatrix} B & C & D\end{pmatrix}^\top \begin{pmatrix} B & C & D\end{pmatrix}= \begin{pmatrix} B^\top B & B^\top C & B^\top D \\ C^\top B & C^\top C & C^\top D \\ D^\top B & D^\top C & D^\top D \\ \end{pmatrix} $$
Consider the following three observations. It seems to me that they lead to a contradiction. Hence, some (all) of them should be wrong. Could you point out the mistake?
(1) If $A$ is invertible, then it does not imply that every square submatrix of $A$ is invertible. For example, we could have that $A$ is invertible but the submatrix $$ Q\equiv \begin{pmatrix} B^\top B & B^\top D \\ D^\top B & D^\top D \\ \end{pmatrix} $$ is not invertible.
(2) $A$ is invertible if and only if $\begin{pmatrix} B & C & D\end{pmatrix}$ is full column rank.
(3) If $\begin{pmatrix} B & D \end{pmatrix}$ is not full column rank, then $\begin{pmatrix} B & C& D \end{pmatrix}$ is not full column rank.
The contradiction is the following:
If $\begin{pmatrix} B & D \end{pmatrix}$ is not full column rank (which holds if and only if $Q$ is invertible), then by (3) $\begin{pmatrix} B & C& D \end{pmatrix}$ is not full column rank. In turn, by (2), $A$ is not invertible. This contradicts (1).
Note. The first version of this answer was written under the misunderstanding that full column rank means that the columns generate the space where they live. The OP has later called my attention to the fact that some people take it to mean that instead, the columns are linearly independent. I must say I find the latter terminology a bit misleading, so I would try to avoid it by refering to it by what it means. Nevertheless, should you like to further investigate the terminology issue, I invite you to compare
https://en.m.wikipedia.org/wiki/Rank_(linear_algebra)
and
https://www.cds.caltech.edu/~murray/amwiki/index.php/FAQ:_What_does_it_mean_for_a_non-square_matrix_to_be_full_rank%3f
Statement 1. Every submatrix of an invertible matrix is invertible.
Verdict: False.
Counter example: Let $A$ be the $3×3$ identity matrix, and consider the $2×2$ submatrix $$ \pmatrix{0 & 1\cr 0 & 0} $$ obtained by removing the first row and the last column of $A$. It is obviously not invertible.
Statement 1 (bis). Every principal submatrix of a positive invertible matrix is invertible.
Verdict: True.
Proof: Let $A$ be a positive invertible $n×n$ matrix. A principal submatrix of $A$ is one that is obtained by removing some rows, say the ones indexed by $$ k_1,k_2,…,k_p, $$ as well as some columns, but not just any columns, rather the columns indexed by the very same $k_1,k_2,…,k_p$.
Letting $R$ be the $n×(n-p)$ matrix whose columns are the canonical basis vectors corresponding to the columns of $A$ which were not removed, the corresponding submatrix may be expressed as $R^TAR$.
In order to prove that $R^TAR$ is invertible, it is enough to prove it is injective, so suppose $x$ is a ($n-p)$-column matrix such that $R^TARx=0$. Then $$ 0 = ⟨R^TARx,x⟩ = ⟨A^{1/2}Rx,A^{1/2}Rx⟩ = \|A^{1/2}Rx\|^2, $$ so $A^{1/2}Rx=0$, and hence $ARx=0$, as well. Since $A$ is injective, then $Rx=0$, but $R$ is also injective, so $x=0$. QED
Statement 2. $A$ is invertible if and only if $\pmatrix{B & C & D}$ has linearly independent columns.
Verdict: True.
Proof: Denoting by the matrix $\pmatrix{B & C & D}$ simply by $E$, suppose that $Ax=0$. Then $$ 0 = ⟨Ax,x⟩ = ⟨E^TEx,x⟩ = ⟨Ex,Ex⟩ = \|Ex\|^2, $$ so $Ex=0$, but since the columns of $E$ are linearly independent, one deduces that $x=0$. QED
Statement 3. If $\pmatrix{B & C}$ does not have linearly independent columns then $\pmatrix{B & C & D}$ does not have linearly independent columns.
Verdict: True.
Proof: Easy exercise.