X is sequentially compact $\implies $ then Lebesgue lemma hold for X where X is metric space
Lebesgue lemma:X is said to satisfy Lebesgue number lemma it for every open cover of X there in $\delta >0$ wherever there is set with diameter $\delta $ then it contains completely in some element of the open cover
I had read proof from Munkres as follows 
I had the following doubts:
1) $x_n $ is constructed in such a way that it implies sequence itself convergent.
I know that by sequentially compactness we get convergent subsequence.
But construction says that $x_n\in B(t,1/n)$ due to this I think there is no choice for sequence rather than to converge. AS we know that it has a convergent subsequence that means sequence converges to the same point
Is my interpretation is correct?
2)I read theorem several times but I do not think I get an idea how to arrive at contradiction.
Please help me to learn this theorem .
Any Help will be appreciated
Best Answer
The non-empty $C_n$ with $\operatorname{diam}(C_n) < \frac1n$ are given separately from each other: the contradiction-assumption only gives they such sets exist (such that $C_n$ is also not a subset of any $A \in \mathcal{A}$) not that we can choose them to be nested, or of the form $B(a,r)$ for some $a,r$ etc. We know nothing at all about them, except their diameter and non-subset properties. That's why we pick $x_n \in C_n$ and use a convergent subsequence of them to get any "grip" on them at all. So we have a convergent subsequence $x_{n_k}$ that converges to some $a \in X$. As $\mathcal{A}$ is a cover, we have some $A_0 \in \mathcal{A}$ that contains this $a$. As $A_0$ is open, there is an $\varepsilon>0$ such that $B(a,\varepsilon) \subseteq A_0$. So far so good.
Now $B(a, \varepsilon)$ has diameter $\le 2\varepsilon$, and it contains infinitely many $x_{n_k}$ (as they converge to $a$) which come from sets of smaller and smaller diameter, so eventually the $C_{n_k}$ they came from are also going to fit inside $B(a,\varepsilon)$: just pick $i$ so large that $\frac{1}{n_i} < \frac{\varepsilon}{2}$ and also such that $d(x_{n_i},a) < \frac{\varepsilon}{2}$, then let $p \in C_{n_i}$. As $p$ and $x_{n_i}$ both comes from $C_{ni}$: $d(p, x_{n_i}) \le \operatorname{diam}(C_{n_i}) < \frac{1}{n_i} < \frac{\varepsilon}{2}$ and so $$d(p,a) \le d(p, x_{n_i}) + d(x_{n_i},a) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$
and as $p \in C_{n_i}$ was arbitrary, $$C_{n_i} \subseteq B(a, \varepsilon) \subseteq A_0$$
which is in immediate contradiction with the fact that no $C_m$ was a subset of any subset of $\mathcal{A}$ by assumption!
The triangle inequality argument was not written down by Munkres (as such arguments are so common that the reader is supposed to fill them in him/herself), but the intuition should be clear: the convergent subsequence forces the small corresponding $C_n$ sets to cluster near $a$ too, and so inside the open $A_0$ that $a$ is in.