Let $U$ be the $N\times N$ unitary matrix that implements a discrete Fourier transform, so that if $v$ is a vector with $N$ complex components (represented by a column matrix), then $Uv$ is its discrete Fourier transform.
Let $\{e_1,...,e_N\}$ be an orthogonal basis in which the components of $U$ are $U_{nm}=\exp(2\pi i\, nm/N)$, let $P$ be a diagonal projection matrix of rank $K$, and define the projected vectors $$ \tilde e_n \equiv U^{-1} P U e_n. $$ This defines a set of $N$ vectors, of which only $K$ are linearly independent. If we choose any $K$ of these $N$ vectors, are they guaranteed to be linearly independent?
(And if the answer is yes, does this phenomenon have a name?)
$U^{-1}$ preserves independence, so we can reformulate your question like this:
Define $PU e_n =: f_n$, and let $V$ be the image of $P$, which is a $k$-dimensional subspace. Is every subset of $\{f_i\}$ of size $k$ a basis for $V$?
Since $V$ was an arbitrary coordinate subspace we can forget about $P$, and just ask:
Does every $k \times k$ minor of the DFT matrix $U$ have a nonzero determinant?
The answer to this is no -- you can find a 2x2 minor of the 4x4 DFT matrix that is the all ones matrix ( https://en.wikipedia.org/wiki/DFT_matrix#Four-point ), and this gives an explicit counter example to your question.
However, it seems that there is a very interesting theorem in the spirit of your question:
Chebotarev's Theorem: The $n \times n$ DFT matrix has the property that all minors are nonzero if and only if $n$ is prime.
See section 1.1 of https://pages.pomona.edu/~sg064747/PAPERS/CNMT.pdf for further discussion, or wikipedia: https://en.wikipedia.org/wiki/Chebotarev_theorem_on_roots_of_unity
One direction of the theorem can be found by generalizing the 4x4 example.