Consider a Toeplitz matrix $T \in \mathbb{R}^{n \times p}$ with randomly independently generated entries and $n < p$. The entries of the Toeplitz matrix are generated by a continuous random variable. Now, let's take $n$ columns $T_1, \dots , T_n$ of $T$.
Are $T_1, \dots, T_n$ linearly independent with probability 1 for any such selection $T_1, \dots , T_n$?
We can prove this in 2 Steps:
Show that each $n \times n$ submatrix $T^\prime$ of $T$ is invertible.
From 1., it follows that each selection of $n$ columns must be linearly independent since they form a basis of $\mathbb{R}^n$.
Proof:
The determinant of every such submatrix $T^\prime$ is a polynomial $p(t_1, \dots, t_j)$, where $t_i$ are the diagonal entries. We know that every polynomial has a finite set of arguments $(t_1, \dots, t_j)$, such that $p(t_1, \dots, t_j)=0$*. Hence, if for all $i$: $t_i\sim \mathbb{P}_i$ for continuous distributions $\mathbb{P}_i$, then the zero-set of $p$ is a set with measure $0$. Therefore, with probability 1 it holds that $\det(T^\prime)=p(t_1, \dots , t_j)\neq 0$, and hence $T^\prime$ must be invertible.
As already explained at the beginning of the answer, we can now derive that the columns of $T^\prime$ form a basis of $\mathbb{R}^n$, and therefore are linearly independent with probability $1$.
*Note that $p\neq 0$.