I know there are some similar questions already on MSE, but none answers my question, since all of them refer to a proof that defines a different function $f(x)$.
Let me first state the version of proof I'm referring to:
We first note that $K(x, \lambda) \subseteq U \iff d(x, U^C) > \lambda$. Therefore $K(x, \lambda) \subseteq U_i$ for at least one $i \iff \text{max}_i \{d(x,U_i^C)\} > \lambda$. Let $U_1, U_2, \ldots, U_n$ be a finite subcover of cover $\mathcal{U}$. Then the function $f: X \to [0, +\infty)$, defined as $f(x) := \text{max}_i \{d(x,U_i^C)\}$ is continuous and has a minimum. This minimum is greater than $0$, since the sets $U_1, \ldots, U_n$ cover $X$. It follows from here that $\lambda := \text{min} \{ f(x); x \in X \}$ is a suitable choice for Lebesgue covering number of the cover $\mathcal{U}$.
My main question is why $f$ is continuous? I certainly do not find it obvious. Then also, why is the minimum strictly positive?
