Triangle Inequality for supremum metric Edited Heavily
Here all functions are from $[0,1]$ to $\mathbb{R}$ and are bounded. 
Prove the following Triangle inequality in following case:
$$\sup_{a\in[0,1]}|X(a)-Y(a)|\le\sup_{a\in[0,1]}|X(a)-Z(a)|+\sup_{a\in[0,1]}|Z(a)-Y(a)|$$
 A: Let $\alpha = \sup_{a \in [0, 1]} |X(a) - Z(a)|$ and $\beta = \sup_{a \in [0, 1]} |Z(a) - Y(a)|$.
$$
\begin{align*}
\alpha + \beta &\ge |X(a) - Z(a)| + |Z(a) - Y(a)| & \text{ for all $a \in [0, 1]$}\\
&\ge |X(a) - Y(a)| & \text{}
\end{align*}$$
Therefore $\alpha + \beta$ is an upper bound on the set $\{|X(a) - Y(a)| : a \in [0, 1]\}$.
Hence by definition of supremum (least upper bound), $\alpha + \beta \ge \sup_{a \in [0, 1]} |X(a) - Y(a)|$.
A: Note I have abused notation below a little, in that I use $x$ both as a 'free' variable and also as the $\sup$ variable. Hopefully it is clear.
A useful result is that if $f(x) \leq g(x)$ for all $x$, then $\sup_x f(x) \leq \sup_x g(x)$. To see this, note that $g(x) \leq \sup_x g(x)$, hence $f(x) \leq \sup_x g(x)$. Taking $\sup_x$ of the left hand side gives $\sup_x f(x) \leq \sup_x g(x)$.
Another useful and obvious result is that if $c$ is a constant, then $\sup_x (f(x) +c) = c + \sup_x f(x)$.
Now consider $f(x)+g(x)$. You have $f(x)+g(x) \leq f(x) + \sup_x g(x)$. Using the above results, we get $\sup_x (f(x) + g(x)) \leq \sup_x (f(x) + \sup_x g(x)) = \sup_x f(x) + \sup_x g(x)$.
Let $f(a) = |X(a)-Y(a)|$, $g(a) = |X(a)-Z(a)|$ and $h(a) = |Z(a)-Y(a)|$. You know that $f(a) \leq g(a)+h(a)$ for all $a$ by the triangle inequality. Now using the above results you have $\sup_a f(a) \leq \sup_a (g(a)+h(a)) \leq \sup_a g(a) + \sup_a h(a)$, which is the result you were trying to prove.
