proof of Lebesgue number lemma

By way of contradiction, suppose that no Lebesgue number existed. Then there exists an open cover $𝒰$ of $X$ such that for all $\delta >0$ there exists an $x\in X$ such that no $U\in 𝒰$ contains $B_{\delta }(x)$ (the open ball of radius $\delta$ around $x$). Specifically, for each $n\in \mathbb{N}$, since $1/n>0$ we can choose an $x_n\in X$ such that no $U\in 𝒰$ contains $B_{1/n}(x_n)$. Now, $X$ is compact so there exists a subsequence $(x_{n_k})$ of the sequence of points $(x_n)$ that converges tosome $y\in X$. Also, $𝒰$ being an open cover of $X$ implies that there exists $\lambda >0$ and $U\in 𝒰$ such that $B_{\lambda }(y)\subseteq U$. Since the sequence $(x_{n_k})$ converges to $y$, for $k$ large enough it is true that ($d$ is the metric on $X$) and . Thus after an application of the triangle inequality, it follows that

$$B_{1/n_k}(x_{n_k})\subseteq B_{\lambda }(y)\subseteq U,$$

contradicting the assumption that no $U\in 𝒰$ contains $B_{1/n}(x_n)$. Hence a Lebesgue number for $𝒰$ does exist.