Show that there exists $\varepsilon >0$ such that for each $x \in X$, the open ball $B(x, \varepsilon)$ is contained in one of the sets in open cover 
Let $\{V_\alpha\}_{\alpha \in A}$ be a finite open cover of a compact metric space $X$. Show that there exists $\varepsilon >0$ such that for each $x \in X$, the open ball $B(x, \varepsilon)$ is contained in one of the $V_\alpha$'s

Since $\{V_\alpha\}_{\alpha \in A}$ is a cover for $X$ we have that $X= \bigcup_{\alpha} V_\alpha.$ This implies that $x \in X$ is also in the union $\bigcup_{\alpha} V_\alpha$. So $x$ is in some of the $V_i$'s. Now for each of $V_i$'s since they're open there exists $\varepsilon_i$ such that $B(x,\varepsilon_i) \subset V_i.$ If I now let $\varepsilon = \min\{\varepsilon_1, \varepsilon_2, \dots, \varepsilon_i\}$ then I have that at least $x \in \bigcup_i V_i$.
I feel like I'm not in the right direction here. I've seen this done by a contradiction, but I'm trying to find what's the problem with this approach. It seems that I've only showed that $x$ will be in the union of the $V_i$'s by taking the minimum of the epsilon's, but $x$ was already in the $V_i$'s by the fact that $x \in\bigcup_{\alpha} V_\alpha$ so of course it would be in the union. What should I do here?
Edit: This is the contradiction argument.
Assume such $\varepsilon$ doesn't exist. Then for every $\varepsilon >0$ there is an $x \in X$ such that $B(x, \varepsilon)$ is not in any $V_\alpha.$ Now for every $n \in \mathbb{N}$ we can choose $x_n \in X$ such that $B(x_n, \frac1n)$ is not in $V_\alpha$ for any $\alpha$. But since $X$ is compact there is a limit point $a \in X$ and since the $V_\alpha$'s for a cover for $X$, then $a \in V_\alpha$ for some $\alpha$. And from here looking at $B(a, r)$ we can find a contradiction.
I'm confused about the fact that we somehow suddenly jump to sequences here in the part "Now for every $n \in \mathbb{N}$ we can choose $x_n \in X$ such that $B(x_n, \frac1n)$ is not in $V_\alpha$ for any $\alpha$" what's going on there?
 A: This is not a valid proof. You need an $\epsilon>0$ which is independent of $x$ but your procedure starts with an $x$ and finds an $\epsilon $ dependent on $x$.
A: The minimum idea can work but you have to be more careful and introduce a new cover to apply compactness too..:
For each $x \in X$ pick $r_x$ such that there is some element $V_{\alpha(x)}$ from $(V_\alpha)_{\alpha \in A}$ and
$$B(x, r_x) \subseteq V_{\alpha(x)}\tag{1}$$
Now, $\{B(x, \frac{r_x}{2})\mid x \in X\}$ is a new open cover for $X$ and so by compactness there is a finite $F \subseteq X$ so that
$$\bigcup \{B(x, \frac{r_x}{2})\mid x \in F\} = X\tag{2}$$
The claim is that $\delta = \min\{\frac{r_x}{2} \mid x \in F\}>0$ is the required Lebesgue number:
Let $x \in X$. Then $x \in B(y, \frac{r_y}{2})$ for some $y \in F$ by $(2)$.
The triangle inequality gives us that $$B(x,\delta) \subseteq B(y, r_y)\tag{3}$$
proof: if $z \in B(x,\delta)$ then $d(x,z) < \delta \le \frac{r_y}{2}$ and $d(x,y) < \frac{r_y}{2}$ too, so $d(y,z) \le d(y,x) + d(x,z) < r_y$, as required for $(3)$.
So from $(1)$ we conclude that $B(x,\delta) \subseteq V_{\alpha(y)}$ and so any $\delta$-ball around any $x$ is contained in some element from the original cover.
A: To address, your query about the contradiction arguement:
Let's first state the hypothesis that we are making for the sake of contradiction: For any $\epsilon > 0$ there exists a point $x_{\epsilon}$ such that $x_{\epsilon}$ is not in any of the open sets in the open cover. In particular we can take $\epsilon$s to be of the form $\frac{1}{n}$ and for each such $\frac{1}{n}$ you can find a point call it $x_n$ such that $x_n$ is not in any $V_{\alpha}$ of the cover. So, this is how you get to the sequence. The motivation for doing is because you are in a second countable space. Basically, the topology can be understood through sequences instead of arbitrary open sets.
Now, the rest is easy. There is a limit of the sequence $x_i$ say $a$.
Now this $a$ is in one of the $V_{\alpha}$ say $a \in V_{m}$. Then you can take a small enough ball for some $\delta > \frac{1}{k}$ for some $k$. But this ball will contain infinitely many points of the sequence. You can take a big enough $n$  such that $B(x_n, \frac{1}{n}) \subset V_{m}$. But u will notice that to ensure this the delta ball around $a$ should be small enough. For this you take a ball of radius $\frac{\delta}{2}$ instead.
So, you have $B(a, \frac{\delta}{2}) \subset B(a, \delta) \subset V_{m} $.
Pick a $x_n \in B(a, \frac{\delta}{2})$ such that $n > 2k$
Check that $B(x_n, \frac{1}{n}) \subset V_m$ which is contradiction to our assumpution!
