In $C[\mathbb{R}]$, if $d(f,g) = \infty$ then $f$ and $g$ are in different connected components.

There's a slight issue with proving that $C=F\cup G$. This is essentially equivalent to the statement you set out trying to prove, namely that everything in the same connected component is a finite distance from one another. Thus while you can prove $C=F\cup G$ (i.e., it is true), any proof of that requires showing that any other $h$ in $C$ is a finite distance from $f$ or a finite distance from $g$, which is only slightly weaker (and not in any useful way, as far as I can tell) that the statement that any $h$ in $C$ is a finite distance from $f$, which is your claim.

Thus you should just directly prove your claim.

The standard approach to this is outlined in the answer to this question.

The idea is that if you have a metric $d: X\to [0,\infty]$, define an equivalence relation on $X$ by $x\sim y \iff d(x,y) < \infty$.

The equivalence classes of $\sim$ are all open, and thus also closed. Hence the connected component of any point $x$ is contained entirely in its equivalence class (which is what you wanted to show).