Proof of Lebesgue Number Lemma for a Sequentially Compact Subset of a Metric Space

Suppose, for contradiction, that no such $\delta$ exists: for every real number $\delta>0$ there is $x\in K$ such that the inclusion $B_d(x,\delta)\subseteq U_i$ fails for every $i\in I$.

By Existence of a Sequence of Positive Real Numbers with Limit Zero fix a sequence $(h_k)_{k\in\mathbb{N}}$ of real numbers with $0<h_k$ for every $k\in\mathbb{N}$ and with limit $0$.

Step 1 (a bad point for each index). For $k\in\mathbb{N}$ put

Each $A_k$ is a subset of $X$, and applying the contradiction hypothesis with $\delta=h_k$ shows $A_k$ is nonempty. By Axiom of Countable Choice, applied to the family of subsets of $X$ $(A_k)_{k\in\mathbb{N}}$, there is a sequence $(x_k)_{k\in\mathbb{N}}$ in $X$ with $x_k\in A_k$ for every $k$; in particular $x_k\in K$.

Step 2 (a convergent subsequence). Since $K$ is sequentially compact, Sequentially Compact Subset of a Metric Space provides $x\in K$ and a strictly increasing sequence $(n_j)_{j\in\mathbb{N}}$ in $\mathbb{N}$, in the sense of Subsequence of a Sequence in a Set, such that $(x_{n_j})_{j\in\mathbb{N}}$ converges to $x$ in $(X,d)$.

Step 3 (a ball around the limit). Since $(U_i)_{i\in I}$ covers $K$ and $x\in K$, there is $i_0\in I$ with $x\in U_{i_0}$. As $U_{i_0}$ is open in $(X,d)$, Open Subset of a Metric Space provides a real number $r>0$ with $B_d(x,r)\subseteq U_{i_0}$. By claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<r\cdot 2^{-1}$ and $r\cdot 2^{-1}+r\cdot 2^{-1}=r$.

Step 4 (choosing an index $j$). By Convergent Sequence in a Metric Space there is $J_1\in\mathbb{N}$ such that $d(x_{n_j},x)<r\cdot 2^{-1}$ for every $j\in\mathbb{N}$ with $J_1\le j$. Since $(h_k)_{k\in\mathbb{N}}$ has limit $0$, Limit of a Sequence of Real Numbers provides $J_2\in\mathbb{N}$ with $|h_k-0|<r\cdot 2^{-1}$ for every $k$ with $J_2\le k$; by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field we have $h_k-0=h_k$, and since $0<h_k$ the absolute value satisfies $|h_k|=h_k$, so $h_k<r\cdot 2^{-1}$ for every such $k$.

By claims 1, 2, and 3 of Properties of the Order on the Natural Numbers, the order on $\mathbb{N}$ is a total order. Let $j$ be the maximum of $J_1$ and $J_2$; by claim 1 of Elementary Properties of the Maximum of Two Elements, $J_1\le j$ and $J_2\le j$. By Strictly Increasing Sequences of Natural Numbers Dominate Their Index, $j\le n_j$, so $J_2\le n_j$ by transitivity. Consequently

Step 5 (contradiction). We claim $B_d(x_{n_j},h_{n_j})\subseteq B_d(x,r)$. Let $y\in B_d(x_{n_j},h_{n_j})$, so $d(x_{n_j},y)<h_{n_j}$ by Open Ball in a Metric Space; with $h_{n_j}<r\cdot 2^{-1}$ and transitivity of $<$, which is claim 2 of Elementary Order Arithmetic in an Ordered Field, this gives $d(x_{n_j},y)<r\cdot 2^{-1}$. By the symmetry axiom of a metric, $d(x,x_{n_j})=d(x_{n_j},x)<r\cdot 2^{-1}$. Adding the two strict inequalities by claim 3 of Elementary Order Arithmetic in an Ordered Field,

and the triangle inequality axiom of a metric gives $d(x,y)\le d(x,x_{n_j})+d(x_{n_j},y)$, so $d(x,y)<r$ by claim 2 of that lemma. Hence $y\in B_d(x,r)$, proving the claim.

Therefore $B_d(x_{n_j},h_{n_j})\subseteq B_d(x,r)\subseteq U_{i_0}$. On the other hand $x_{n_j}\in A_{n_j}$, so by the definition of $A_{n_j}$ the inclusion $B_d(x_{n_j},h_{n_j})\subseteq U_i$ fails for every $i\in I$, and in particular for $i=i_0$. This contradiction proves the lemma.