Yes, it is. First we prove that if , where 's are linearly independent, then each too. Let . Let , then . We can write this relation in matrix: . Since is linearly independent, so matrix is non-singular, then we have , that is, each can be expressed as a linear combination of with coefficients from . Because , so is and .
Now if the distance of and , i.e. , then we have for each according to the preceding paragraph, this just means that the distance of and is arbitrarily small.
If we are discussing in where , then the linear independence of implies that has determinantal rank , so we can shrink to be the -rowed minor whose det 0 and the rest discussion remains unchanged.
Generally, I don't like to handle the low-level elements, so if there is some other better proof, please follow, thanks.