One approach of arguable snazziness is as follows. The $\implies$ direction is straightforward. For the $\Longleftarrow$ direction, suppose without loss of generality that $X$ is symmetric and that $\tilde X$ is the leading $r \times r$ principal submatrix. We have
$$
X = \pmatrix{\tilde X&B\\B^T&C}.
$$
We note that $X$ is positive semidefinite if both $\tilde X$ and $X/\tilde X$ are positive semidefinite, where $X/\tilde X$ is the Schur complement. That is,
$$
X/\tilde X = C - B^T \tilde X^{-1}B.
$$
On the other hand, the rank of $X$ must be $r$, so we have
$$
r = \operatorname{rank}(X) = \operatorname{rank}(\tilde X) + \operatorname{rank}(X/\tilde X) = r + \operatorname{rank}(X/\tilde X).
$$
Thus, $X/\tilde X = 0$, which means that $X/\tilde X$ is positive semidefinite. We conclude that $X$ is indeed positive semidefinite.
The $\implies$ direction in detail: suppose without loss of generality that the first $r$ columns of $X$ form the maximal linearly independent set in question. That is, $Xv \neq 0$ for all $v$ in the span of $\{e_1,\dots,e_r\}$ (where $e_j$ is the $j$th canonical basis vector). It follows that $v^TXv \neq 0$ for all $v$ in the span of $\{e_1,\dots,e_r\}$. From this, we conclude that the leading $r \times r$ submatrix $\tilde X$ is positive definite, as desired.
