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). The subset $K$ is compact in $(X,\mathcal{T}_d)$, so statement 1 of Compact Subset Criterion via Open Covers in the Ambient Space holds for $K$, and therefore so does statement 2 of that theorem. Applied to the open cover of Step 1, whose index set is $X$, it yields a finite subset $F\subseteq X$ with

$$K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon).$$

Since $\varepsilon>0$ was arbitrary, and $F$ is a finite subset of $X$ whose balls of radius $\varepsilon$ cover $K$, Totally Bounded Subset of a Metric Space shows that $K$ is totally bounded in $(X,d)$.