Proving that $d(f,g)=\|f-g\| = \sup \limits_{0\leq x \leq 1} |f(x)-g(x)|$ is a metric on $X=C_b[0,1]$ Following Proving that $d(f,g)=\|f-g\| = \sup \limits_{0\leq x \leq 1} |f(x)-g(x)|$ is a metric on $X=C[0,1]$ 
I would like to prove that the same is true for bounded functions on $[0,1]$ ,$C_b([0,1]) = \{f:X \to Y| f \text{ bounded }\}$ (not necessarily continuous), however I can't see where I could use the boundedness property. I am producing a proof that is essentially identical to the linked and that confuses me a little bit.
Proof attempt: Obvious  $d(f,g) = \sup \limits_{0\leq x \leq 1} |f(x)-g(x)| \geq 0 $ and $d(f,g) = d(g,f)$. 
Triangle inequality: 
Let,$f,g,h$ be bounded functions, then 
$f(x) - h(x) = f(x) - g(x) + g(x) - h(x) \leq |f(x) - g(x)| + |g(x) - h(x)| \leq \sup \limits_{0\leq x \leq 1} |f(x)-g(x)| +  \sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$
Then taking $\sup$ over LHS yields $\sup \limits_{0\leq x \leq 1} |f(x)-h(x)| \leq\sup \limits_{0\leq x \leq 1} |f(x)-g(x)| +  \sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$
But no where in above proof did I use boundedness assumption. Can someone please point out where an adjustment could be made?
Thanks

You need boundedness to be able to define the metric. 
On the notation side, usually $C_b$ is reserved for "continuous and bounded". A more common notation (although not universal) would be $B(X)$. 
