In my textbook I have following definition for the sup-Norm of bounded functions on an arbitrary set $X$ taking values in $\mathbb{C}$, i.e. $B(X) := \{\, f \colon X \to \mathbb{C} \mid \sup_{x \in X} |f(x)| < \infty \,\}$:
$$|| \cdot ||_{\infty} \colon B(X) \ni f \mapsto \sup_{x \in X} |f(x)| \in \mathbb{R}_{\geq 0} \qquad (\ast)$$ Later in the textbook there occurs the definition of $|| \cdot ||_{\infty}$ for a more general situation again, i.e. if $(Y,|| \cdot ||)$ is a normed space and we define $B(X,Y) := \{\, f \colon X \to Y \mid \sup \{\,d_{||\cdot||}(f(x),f(x^\prime) \mid x,x^\prime \in X\,\} < \infty \,\}$, then author puts
$$|| \cdot ||_{\infty} \colon B(X,Y) \ni f \mapsto \sup \{\, d_{||\cdot||}f(x) ,f(x^\prime)) \mid x,x^\prime \in X\,\}\in \mathbb{R}_{\geq 0} \qquad (\ast \ast)$$
where $d_{||\cdot||}(f(x),f(x^\prime)) := ||f(x) - f(x^\prime)||$, i.e. metric induced by the norm $||\cdot||$.
If I choose $(Y,||\cdot||) = (\mathbb{C},|\cdot|)$, why are equations $(\ast)$ and $(\ast \ast)$ equivalent?
PS: Since the author uses the same symbols for notation, I assume that both equations need to be equal and furthermore the special case shall be included in the more general one.
No, they're not the same, but by the triangle inequality (via $0$) $\|f\|^{(2)}_\infty \le 2\|f\|^{(1)}_\infty$, where the superscript denotes the first or the second definition. So for most purposes (like (uniform) continuity, etc.) these give the same result.