Q:

Understanding a step in The Lebesgue Number Lemma

Lemma: Let $\mathcal{A}$ be an open covering of the metric space
  $(X,d)$. If $X$ is compact, then there is $\delta >0$ such that for
  each subset of $X$ having diameter less than $\delta$, there exists an
  element of $\mathcal{A}$ containing it.

Proof: 
Let $\mathcal{A}$ be an open covering of $X$. IF $X$ itself is an element of $\mathcal{A}$ then we can let $\delta$ to be any positive number. 
Second Case: $X \notin \mathcal{A}$. Since $X$ is compact, $\mathcal{A}$ has a finite subcollection say, $A_1 ,A_2, ...,A_n$ that covers $X$. Now define :
$$f(x) : = \frac{1}{n} \sum_{i=1}^{n} d(x,C_i)$$
where $C_i = X - A_i$ for $1 \le i \le n$. 
First, we need to show that $f>0$. given $x \in X$ choose $i$ so that $x \in A_i$. Then choose $\epsilon$ such that the $\epsilon$- neighoubrhood of $x$ lies in $A_i$. Then $d(x_i,C_i)> \epsilon,$ so that $f(x)> \frac{\epsilon}{n}$, so $f$ becomes positive.
After we show that since $f$ is continuous it attains a minimum. This minimum would be our specified lebesgue number. ($\delta$).
I have no problems with the last step, but in the step in bold I have a problem. I know that we need to show that $f$ is $>0$, but isn't it obvious that $f$ is $>0$. By definition:
$$d(x,C_i) : = \inf(d(x,c) \ \text{where} \ c \in C_i)$$
and since $d$ is a metric by the properties of metrics $d(x,y) \geq 0$ $\forall x,y$. Then it follows easily that $f\geq0$. Since $A_i$ cover $X$, then $$\bigcap C_i = \bigcap (X- A_i) = X - \bigcup (A_i) = \emptyset$$.
then $f$ can not be $0$. So $f>0$. 
Q: Is what I'm doing right? Is this another way of showing that $f>0$? Isn't this way more clear? Why would anybody use the way above instead? (This proof is from Munkres' book on Topology.) 
Thanks.

A:

It would be safer to explicitly state that $x \notin C_i \implies d(x,C_i) > 0$ since the $C_i$ are closed. Without that, $\bigcap C_i = \varnothing$ doesn't imply that $f$ is strictly positive. But keeping this implicit is the only (minor) issue with your argument.
However, this isn't much different from Munkres' argument. The difference is that you are using facts (that probably have been proven before) which Munkres explicitly spells out (again). Munkres shows, as part of his argument, that $d(x,C_i) > 0$ for $x \notin C_i$ since $C_i$ is closed, which you tacitly use.
It is of course more efficient to re-use results, rather than prove them more than once, but that may be pædagogically suboptimal. Some redundancy, including reproving results, helps people learn. (The amount of redundancy where it changes from being helpful to being bothersome varies from person to person of course.)

