Proof of A Compact Subset of a Metric Space is Totally Bounded

Let $\varepsilon$ be a real number with $\varepsilon>0$.

Step 1 (the balls of radius $\varepsilon$ cover $K$). For $x\in X$ let $B_d(x,\varepsilon)$ be the open ball with center $x$ and radius $\varepsilon$, which is open in $(X,d)$ by Open Ball in a Metric Space is Open and so lies in $\mathcal{T}_d$. Regard $(B_d(x,\varepsilon))_{x\in X}$ as a family of subsets of $X$ indexed by $X$. If $y\in K$, then $y\in X$ and $d(y,y)=0<\varepsilon$ by the identity-of-indiscernibles axiom of a metric, so $y\in B_d(y,\varepsilon)$. Hence this family is an open cover of $K$ in $(X,\mathcal{T}_d)$.

Step 2 (a finite subcover). Since $K$ is compact in $(X,\mathcal{T}_d)$, Compact Subset Criterion via Open Covers in the Ambient Space yields a natural number $n$ and points $a_1,\dots,a_n\in X$ indexed by the initial segment $[n]$, forming a tuple in $X$, such that

$$K\subseteq B_d(a_1,\varepsilon)\cup\cdots\cup B_d(a_n,\varepsilon).$$

Step 3 (the set of centers is finite). Let $F=\{a_i:i\in[n]\}\subseteq X$. The assignment $i\mapsto a_i$ is a surjection from $[n]$ onto $F$, and $[n]$ has $n$ elements by claim 1 of Basic Properties of Finite Sets; hence $F$ is finite by claim 4 of that lemma. Moreover

$$\bigcup_{a\in F}B_d(a,\varepsilon)=B_d(a_1,\varepsilon)\cup\cdots\cup B_d(a_n,\varepsilon),$$

since the two unions are taken over the same collection of balls. Combining with Step 2 gives $K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon)$.

Since $\varepsilon>0$ was arbitrary, Totally Bounded Subset of a Metric Space shows that $K$ is totally bounded in $(X,d)$.