I'm questioning my logic on this question and want to verify that my reasoning is correct.
Let $S\neq\emptyset$ and $X=\{\varphi:S\rightarrow\mathbb{R}|\; \varphi \text{ is bounded} \}$. Define $D:X\times X\rightarrow \mathbb{R}$ by $$D(\varphi_1,\varphi_2) = \sup_{u\in S}|\varphi_1(u)-\varphi_2(u)|.$$ Because $\varphi_1,\varphi_2$ are bounded, $D$ is well-defined and $(X,D)$ is a metric space, hence a topological space with the induced metric topology.
If $X$ is compact, then $d:S\times S\rightarrow\mathbb{R}$ defined by $$ d(u,v) = \sup_{\varphi\in X}|\varphi(u)-\varphi(v)| $$ is well-defined and is a pseudo-metric on $S$.
Notice that $$d(u,v)=0 \Leftrightarrow \varphi(u)=\varphi(v) \;\;\; \forall\phi\in X.$$
In general, this does not imply that $u=v$. However, if $|S|\leq|\mathbb{R}|$, then there exists a bounded, injective $\psi:S\rightarrow\mathbb{R}$ because the cardinality of $S$ is the same as $[-1,1]$. So because $\psi$ is injective, $u=v$.
This means that as long as $|S|\leq|\mathbb{R}|$, then $d$ is a metric.
Is my logic sound here?
Well, what you say is correct, but only vacuously so: $X$ is never compact. Indeed, for any $r\in\mathbb{R}$, let $\varphi_r$ be the constant function with value $r$. Then $\varphi_r\in X$ for all $r$, and $D(\varphi_r,\varphi_s)=|r-s|$. In particular, $X$ is unbounded, so it cannot be compact.
Note moreover that (ignoring the requirement that $X$ be compact), $d(u,v)$ is always infinite if $u\neq v$ (and so $d$ is never a (pseudo)metric if $|S|>1$, unless you allow (pseudo)metrics to take the value $\infty$). For instance, you can consider functions $\varphi_{u,r}$ defined by $\varphi_{u,r}(u)=r$ and $\varphi_{u,r}(x)=0$ for all $x\neq u$ for each $r\in\mathbb{R}$.