Proving the distance function $|f - g|_u = \sup \{ |f(x) - g(x)|: x \in S \}$ defines a metric space Let $S$ be a closed and bounded subset of $\mathbb{R}$. Define the "functional distance" between $f$ and $g$, both functions from $S$ to $\mathbb{R}$, to be
$$
|f - g|_u = \sup \{ |f(x) - g(x)|: x \in S \}.
$$
Show that this makes the set of continuous functions from $S$ to $\mathbb{R}$ into a metric space.

I'm confused on how to approach this problem. Do I just need to show that
$$
|f - g|_u = \sup \{ |f(x) - g(x)|: x \in S \}.
$$
satisfies the definition of a metric given in my textbook or is there something deeper I should be thinking about? Seems that if I show that this is a valid metric, the claim follows immediately by the definition of a metric space.
 A: In this answer $d\left(f,g\right)$ is an alternative notation for
$\left|f-g\right|_{u}$. 
Define $\left\Vert f\right\Vert :=\sup\left\{ \left|f\left(s\right)\right|\mid s\in S\right\} $ and note that $d\left(f,g\right)=\left\Vert f-g\right\Vert $.
When it comes to the triangular inequality then it is enough to prove
that $\left\Vert f+g\right\Vert \leq\left\Vert f\right\Vert +\left\Vert g\right\Vert $
because this leads automatically to: $$d\left(f,h\right)=\left\Vert f-h\right\Vert =\left\Vert \left(f-g\right)+\left(g-h\right)\right\Vert \leq\left\Vert f-g\right\Vert +\left\Vert g-h\right\Vert =d\left(f,g\right)+d\left(g,h\right)$$
It is to be shown that: $$\sup\left\{ \left|f\left(s\right)+g\left(s\right)\right|\mid s\in S\right\} \leq\sup\left\{ \left|f\left(s\right)\right|\mid s\in S\right\} +\sup\left\{ \left|g\left(s\right)\right|\mid s\in S\right\} $$
That is an immediate consequence of:
$\left|f\left(t\right)+g\left(t\right)\right|\leq\left|f\left(t\right)\right|+\left|g\left(t\right)\right|\leq\sup\left\{ \left|f\left(s\right)\right|\mid s\in S\right\} +\sup\left\{ \left|g\left(s\right)\right|\mid s\in S\right\} $
for each $t\in S$.
