Let $X$ be a nonempty set and $(Y,d)$ be a metric space, define $$B(X,Y) := \{\, f \colon X \to Y \mid \sup \{\, d(f(x),f(x^\prime))\mid x.x^\prime \in X \,\} < \infty \,\} $$ and $$ d_\infty \colon B(X,Y) \times B(X,Y) \ni (f,g) \mapsto \sup \{\, d(f(x),g(x) \mid x \in X \,\} \in \mathbb{R}_{\geq 0}$$.
Assume we have shown that $(B(X,Y),d_\infty)$ is a metric space. In the case that $(Y,||\cdot||_\infty)$ is a normed space, then $d_\infty$ as a metric is induced by the norm $$ ||\cdot||_\infty \colon B(X,Y) \ni f \mapsto \sup \{\, ||f(x) -f(x^\prime)|| \mid x,x^\prime \in X \,\} \in \mathbb{R}_{\geq 0}$$.
It is straightforward to show that $||\cdot||_\infty$ defines a norm on $B(X,Y)$. But it I fail to show that $||\cdot||_\infty$ induces $d_\infty$. Since $$ || f - g|| = \sup \{ ||(f-g)(x) - (f-g)(x^\prime)|| \mid x,x^\prime \in X \,\},$$ but how this should be equal to $$ \sup \{ || f(x) - g(x) || \mid x \in X \,\} = d_\infty(f,g)?$$
There's probably something wrong with your question. Have a look at this example:
Let $X=\mathbb{R}$, $(Y, \|\cdot\|) = (\mathbb{R}, |\cdot|)$, the Euclidean line and let $f(x)=\sin(x)$. Obviously $f\in B(X,Y)$. It is $\|f\|_\infty=\sup\{|\sin(x)-\sin(x')|: x,x'\in\mathbb{R}\}=2$. Hence $\|f-0\|_\infty=2$. But on the other hand, $d_\infty(f,0)=\sup\{|\sin(x)|: x\in\mathbb{R}\}=1$.