Supremum norm for the set of continuous functions 
Let $D \ne \emptyset$ and $(X,d)$ be a metric space. Let $Z =(D,X)$, that is the set of all $\textbf{bounded}$ continuous functions $f:D \to X$. Now $Z$ isn't a vector space, but we can  define the sup metric  $e$ for $Z$ by setting $$e(f,g) = \sup_{x\in D} d(f(x),d(g(x)).$$

This was a definition on a book I'm reading without a proof and I'm trying to figure out if this satisfies the conditions for a metric space. If it does it should satisfy $$d(x,y) \geqslant 0 \text{ with equality iff $x=y$} \\ d(x,y) = d(y,x) \\ d(x,y) \leqslant d(x,z) + d(z,y)$$
I'm slightly confused about how to show these. For the first one I would want to show that $\sup_{x\in D} d(f(x),d(g(x)) \geqslant 0$, which seems intuitive that the distance between two functions wouldn't be negative, but I'm not sure how to formalize this. Also the second one seems self explanatory that the distance between $f$ and $g$ is the same as the distance between $g$ and $f$. If anyone happens to know where this would be proven I would be appreciate a link.
 A: Your arguments are superficial. You say for example that symetry is plausible because we are talking about a distance. Don't forget: you are trying to prove that $e$ is a metric, don't use it as an assumption!
For instance, the first one goes as follows: for every $x\in D$, we have $d(f(x),f(g))\geq 0$, because $d$ is a metric on $X$. The supremum of a set of non-negative numbers is non-negative (or infinity...). Moreover, it is zero iff all numbers in the set whose supremum is considered are 0. So indeed, $e(f,g)\geq 0$ (including the possibility that it is infinite). Equality holds iff $d(f(x),f(y))=0$ for all $x\in D$, that is, iff $f(x)=g(x)$ for all $x\in D$, that is, iff $f=g$.
As @Kavi Rama Murthy  already pointed out, the statement itself is false in general, because of the possibility of infinite distances. This is usually fixed by restricting the attention to bounded (continuous) functions. For example, if $D$ is compact, then continuous functions are bounded, so the finiteness of the supremum metric is guaranteed.
A: You just have the apply the fact that $d$ is a metric, plus the definition of sup.
You don't need continuity of the functions, just their boundedness.
As $f$ and $g$ are bounded, this means that there are $M,N \in \Bbb R$ such that $f[D]$ is contained in some ball $B(x_1,M)$ and $g[D]$ is contained in some ball $B(x_2,N)$, say.
But then for all $x \in D$ we have $$d(f(x), g(x)) \le d(f(x), x_1) + d(x_1, x_2) + d(x_2, g(x)) < M+ d(x_1,x_2) + N$$
so that the set $\{d(f(x), g(x)\mid x \in D\}$ is bounded above in $\Bbb R$ (the right hand side is fixed and does not depend on $x$).
So $e(f,g)$ is well-defined, as in $\Bbb R$ every set that is bounded above has a supremum in $\Bbb R$. For any fixed $x\in D$ we have $0 \le d(f(x), g(x)) \le e(f,g)$ (as a sup is an upperbound for all values and $d$ is a metric) so $e(f,g) \ge 0$ follows immediately.
Suppose that $f \neq g$. Then for some $x \in D$ we have that $f(x) \neq g(x)$; this is what equality of functions means. But then for that $x$, $0 < d(f(x), g(x)) \le e(f,g)$ as $d$ is a metric and so $e(f,g)>0$ for distinct functions, which is what we had to prove (in contrapositive form).
Now if $f,g \in Z$ for all $x$ we have $d(f(x), g(x)) = d(g(x), f(x))$ as $d$ is a metric. So the sets $\{d(f(x), g(x)) \mid x \in X\}$ and $\{d(g(x), f(x)) \mid x \in X\}$ are exactly the same set, so their sups, resp $e(f,g)$ and $e(g,f)$ are the same too. This shows symmetry.
Finally, for the triangle inequality, let $f,g,h \in Z$.
For any fixed $x \in D$ we have
$$d(f(x), h(x)) \le d(f(x), g(x)) + d(g(x), h(x)) \le e(f,g) + e(g,h)$$
using that $d$ obeys the triangle inequality and the sups are upperbounds. So the real number $e(f,g)+e(g,h)$ is an upperbound for the set $\{d(f(x), h(x)) \mid x \in X\}$ and $e(f,h)$ is the smallest upperbound of that same set so $e(f,h) \le e(f,g)+e(g,h)$ follows and we're done.
