Proof of Every Sequence in a Compact Subset of a Metric Space Has a Cluster Point There

Suppose, for contradiction, that no point of $K$ is a cluster point of $(x_m)_{m\in\mathbb{N}}$.

Step 1 (a family of open sets that the sequence eventually avoids). Let

We regard $\mathcal{U}$ as a family of subsets of $X$ indexed by $\mathcal{U}$ itself, the member indexed by $U$ being $U$.

Step 2 ($\mathcal{U}$ covers $K$). Let $x\in K$. By assumption $x$ is not a cluster point of $(x_m)$, so by Cluster Point of a Sequence in a Metric Space there are a real number $\varepsilon>0$ and $N\in\mathbb{N}$ such that no $m\in\mathbb{N}$ with $N\le m$ satisfies $d(x_m,x)<\varepsilon$. Consider the open ball $B_d(x,\varepsilon)$, which is open in $(X,d)$ by Open Ball in a Metric Space is Open, hence lies in $\mathcal{T}_d$. If $m\in\mathbb{N}$ satisfies $N\le m$, then $d(x_m,x)<\varepsilon$ fails, and by the symmetry axiom of a metric $d(x,x_m)=d(x_m,x)$, so $d(x,x_m)<\varepsilon$ fails and therefore $x_m\notin B_d(x,\varepsilon)$. Hence $B_d(x,\varepsilon)\in\mathcal{U}$. Moreover $d(x,x)=0<\varepsilon$ by the identity-of-indiscernibles axiom of a metric, so $x\in B_d(x,\varepsilon)$. Since $x\in K$ was arbitrary, $K$ is contained in the union of the members of $\mathcal{U}$, so $\mathcal{U}$ is an open cover of $K$ in $(X,\mathcal{T}_d)$.

Step 3 (a finite subcover). Since $K$ is compact in $(X,\mathcal{T}_d)$, Compact Subset Criterion via Open Covers in the Ambient Space yields $n\in\mathbb{N}$ and members $U_1,\dots,U_n$ of $\mathcal{U}$ with

Write $[n]$ for the initial segment determined by $n$, which indexes this finite list.

Step 4 (a common threshold). For $U\in\mathcal{U}$ let

which is nonempty by the definition of $\mathcal{U}$, and set $N_U=\min W_U$, which exists by The Natural Numbers Are Well Ordered. This defines the tuple $(N_{U_i})_{i\in[n]}$ in $\mathbb{N}$, with no appeal to a choice principle. By claims 1, 2, and 3 of Properties of the Order on the Natural Numbers, the order $\le$ on $\mathbb{N}$ is reflexive, transitive, antisymmetric, and any two natural numbers are comparable, so it is a total order. Hence Greatest Element of a Finite Family in a Totally Ordered Set provides $j\in[n]$ with $N_{U_i}\le N_{U_j}$ for every $i\in[n]$. Put $N=N_{U_j}$.

Step 5 (contradiction). Consider the index $m=N$, which satisfies $N\le m$ by reflexivity of $\le$. By hypothesis $x_N\in K$, so by Step 3 there is $i\in[n]$ with $x_N\in U_i$. On the other hand $N_{U_i}\le N$ by Step 4, and $N_{U_i}\in W_{U_i}$, so applying the defining property of $W_{U_i}$ with the index $N$, which satisfies $N_{U_i}\le N$, gives $x_N\notin U_i$. This is a contradiction.

Therefore the assumption was false, and some $x\in K$ is a cluster point of $(x_m)_{m\in\mathbb{N}}$ in $(X,d)$.