Prove the existence of a positive number $c$ This question is from the third edition of Bert Mendelson's "Introduction to Topology". The following is exercise number four of section five, chapter five (on page 178).
Problem Statement: Let $(X,d)$ be a compact metric space and let $\{F_{\alpha}\}_{\alpha \in I}$ be a family of closed subsets of $X$ such that $\bigcap_{\alpha \in I}F_{\alpha} = \varnothing$. Prove that there is a positive number $c$ such that for each $x \in X$, $d(x,F_{\alpha}) \geq c$ for some $\alpha \in I$.
Some useful theorems and definitions (from the aforementioned book):
Definition of Lebesgue number: Given an open covering $\{O_{\alpha}\}_{\alpha \in I}$ of a metric space $(X,d)$, if a positive number $\epsilon$ has the property that for each $x \in X$, $B(x;\epsilon) \subset O_{\beta}$ for some $\beta \in I$, then each $\epsilon'$ with $0<\epsilon'<\epsilon$ also has the same property. The least upper bound of the set of numbers having this property is called the Lebesgue number, $\epsilon_L$, of the open covering $\{O_{\alpha}\}_{\alpha \in I}$.
Corollary 5.6: Let $(X,d)$ be a metric space such that each infinite subset of $X$ has an accumulation point. Then each open covering $\{O_{\alpha}\}_{\alpha \in I}$ of $X$ has a Lebesgue number $\epsilon_L$.
Theorem 5.8: Let $(X,d)$ be a metric space. Each infinite subset of $X$ has at least one accumulation point if and only if $X$ is compact.
My Attempt: Set $O_{\alpha} = X-F_{\alpha}$. Then from the hypothesis, $\{O_{\alpha}\}_{\alpha \in I}$ covers $X$ and is a family of open sets. Corollary $5.6$ along with Theorem $5.8$ implies that there is a Lebesgue number $\epsilon_L$ for this family of open sets. Set $c=\epsilon_L$. Then for each $x \in X$, $B(x;c) \subset O_{\beta}$ for some $\beta \in I$ since our $c$ is equal to the Lebesgue number $\epsilon_L$. For this choice of $\beta$, $d(x,F_{\beta}) \geq c$ as otherwise, $d(x,F_{\beta}) < c$ implies that there exists $x' \in F_{\beta}$ such that $d(x,x') < c$ which in turn implies (as $B(x;c) \subset O_{\beta}$) that $x' \in O_{\beta}$, a contradiction as these two sets are complements of each other by definition.
Therefore this positive number $c = \epsilon_L$ is such that for each $x \in X$, $d(x,F_{\alpha}) \geq c$ for some $\alpha \in I$.
Admittedly, I hand-waived the last part of the proof where I concluded the existence of $x' \in F_{\beta}$ such that $d(x,x') < c$ if we assumed $d(x,F_{\beta}) < c$. This must be a simple exercise in the $\epsilon - \delta$ world and I simply assumed it to be true.
Please let me know if my proof looks alright.
 A: Your proof is completely fine. The part you are worried about can be considered obvious, but if needed, the elaboration goes like this:
Assume $d(x, F_{\beta}) < c$. Since by definition $d(x, F_{\beta})$ is the greatest lower bound of the set $\{ d(x, x') : x' \in F_{\beta} \}$, the number $c$ can not be a lower bound of this set. This means that $d(x, x') < c$ for some $x' \in F_{\beta}$, which is what we need.
A: Here is a sketch of a possible alternative proof.
Assume towards a contradiction that there were no such positive number $c$.
Then for every $c>0$ there is $x$ (depending on $c$) such that
$d(x,F_\alpha)<c$ for all $\alpha \in I.$ (This is just an exercise in negating the conclusion of the Problem.)
In particular, for every positive integer $n$ (taking $c=\dfrac1n$) there is $x_n$ such that $d(x_n,F_\alpha)<\dfrac1n$ for all $\alpha \in I.$ Now, since $X$ is compact metric, the sequence $x_n$ must have a converging subsequence, say $x_{n_k}$ converging to some $y$ as $k\to\infty.$ Show that $y\in\bigcap_{\alpha \in I}F_{\alpha},$ thus obtaining a contradiction. Indeed, take any $\varepsilon>0$ and show that $d(y,F_\alpha)<\varepsilon$ for all $\alpha.$ To do this, take $k$ large enough so that:
(a) $\dfrac1{n_k}<\dfrac\varepsilon2$ and
(b) $d(x_{n_k},y)<\dfrac\varepsilon2.$
Then $d(y,F_\alpha)\le d(y,x_{n_k})+d(x_{n_k},F_\alpha)<\dfrac\varepsilon2+\dfrac1{n_k}<\dfrac\varepsilon2+\dfrac\varepsilon2=\varepsilon.$
Letting $\varepsilon\to0,$ conclude that $d(y,F_\alpha)=0$ for all $\alpha$ , and since the $F_\alpha$ are closed it follows that $y\in F_\alpha$ for all $\alpha\in I$, a contradiction.
