Here, we have $F(k,n)$ defined as the set of ordered $k$-tuples of linearly independent vectors in $\mathbb{R^n}$.
To start, let $X \in F(k,n)$. We can express $X$ as
$$X = \begin{bmatrix} x_1^1 & \cdots & x_1^n \\ \vdots & \vdots & \vdots \\ x_k^1 & \cdots & x_k^n \end{bmatrix}$$
where this is a matrix of rank $k$. Having rank $k$ implies that not all $k \times k$ minors are equal to zero. Define a function:
$$f(X) = \sum(k\times k \text{ minors of } X)^2$$ This is a continuous function from $\mathbb{R}^{k\cdot n} \to \mathbb{R}$.
Using definition of continuity, we get that $f^{-1}(\{0\})$ is a closed subset of $\mathbb{R}^{n\cdot k}$.
My question is, why does $f^{-1}(\{0\})$ being closed imply that $F(k,n)$ is an open subset of $\mathbb{R}^{k\cdot n}$?
Yes. The complement of a closed set is open. Alternately, $$F(k,n)= f^{-1}( (- \infty,0) \cup (0, \infty)$$ is the pre-image of an open set under a continuous function.