Probability that $n$ vectors drawn from arbitrary independent prob. distributions are linearly independent

Let $X_1, X_2, \ldots, X_n \in \mathbb{R}^n$ be continuous random variables in $\mathbb{R}^n$ that are pairwise and elementwise independent. By continuous random variables, it is meant that their distributions do not include any Dirac delta fnc $\delta_x$.

My question is:

in this general situation, are $X_1, \ldots, X_n$ linearly independent (LI) with propbability 1? That is, is it true that $$\operatorname{Prob}\big ( \operatorname{span} \{ X_1, \ldots, X_n\} = \mathbb{R}^n \big ) = 1\text{?}$$

This is a general version of the topic:

Probability that $n$ vectors drawn randomly from $\mathbb{R}^n$ are linearly independent

since here we consider a general independent prob. distribution, rather than uniform (or Gaussian) distribution. By referring the related posts above and below:

The probability that two vectors are linearly independent.

I think the statement above is true as shown in the short proof below.

First, the probability of "$X_1 = 0$" is zero since the singleton $\{ 0 \}$ is a set of measure zero. Hence, $X_1$ is LI with probability 1. Next, assume that $X_1,\ldots,X_r$ are LI with probability 1 for some $r \in \{1,2,\ldots, n-1\}$. Then, with probability 1, they span an $r$-dimensional subspace of $\mathbb{R}^n$ which is, however, a set of measure zero, too. Hence, the next vector $X_{r+1}$ lies on $\mathbb{R}^n$ outside the subspace with prob. $1$. By this process, $\operatorname{span}\{X_1,\ldots,X_n\} = \mathbb{R}^n$ with prob. $1$.

But, I'm just not sure that this proof is correct and has no problem.