Lebesgue's Number Lemma

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$M$$ be sequentially compact.

Then there exists a Lebesgue number for every open cover of $$M$$.

Proof
Suppose that $$\mathcal{U}$$ is an open cover of $$M$$ for which no Lebesgue number exists.

Then for any $$n \in \N$$, there exists some $$x_n \in M$$ such that $$N_{1/n} \left({x_n}\right) \subseteq U$$ is false for each $$U \in \mathcal{U}$$. (Otherwise $$1/n$$ would be a Lebesgue number for $$\mathcal{U}$$.)

As $$M$$ is sequentially compact, $$\left \langle {x_n} \right \rangle$$ has a subsequence, say $$\left \langle {x_{n \left({r}\right)}} \right \rangle$$ which converges to some $$x \in M$$.

Since $$\mathcal{U}$$ covers $$M$$, $$x \in U_0$$ for some $$U_0 \in \mathcal{U}$$.

Since $$U_0$$ is open, $$\exists m \in \N: N_{2/m} \left({x}\right) \subseteq U_0$$.

Now $$N_{1/n} \left({x}\right)$$ contains $$x_{n \left({r}\right)}$$ for all $$r \ge R$$, say.

Choose $$r \ge R$$ such that $$n \left({r}\right) \ge m$$ and write $$s = n \left({r}\right)$$.

Then $$N_{1/s} \left({x_s}\right) \subseteq N_{2/m} \left({x}\right)$$ since:

$$ $$ $$ $$

So $$N_{1/s} \left({x_s}\right) \subseteq U_0$$ which contradicts the choice of $$x_s$$.

So there has to be a Lebesgue number for $$\mathcal{U}$$.