Let $\epsilon_0$ be the fixed positive constant such that $F_0 := \{ y \mid b(y) < \epsilon_0\}$ is nonempty and bounded. Fix $z\in F_0$. Then we have that if $y \in Y \setminus F_0$, $b(y) - b(z) \geq \epsilon_0 - b(z) =: \delta_0 > 0$.
By assumption $F_0$ is bounded, hence there exists $C > 0$ such that $B_C(z) \supset F_0$. (Here I take $B_C(z)$ to be the open ball; a simple modification of the following argument also allows for closed balls.)
Let $c = \frac{\delta_0}{2C}$. Now we verify this choice of $C$ and $c$ is good.
Given $y \in Y\setminus B_C(z)$, let $\tilde{y} = \frac{C}{\|y-z\|} (y-z) + z$. By convexity of the function $b$ we have that
$ \frac{C}{\|y-z\|} (b(y) - b(z)) + b(z) \geq b(\tilde{y}) \geq b(z) + \delta_0 $
Hence we have that
$ b(y) - b(z) \geq \frac{\delta_0}{C} \|y-z\| > \frac{\delta_0}{2C} \|y-z\| = c \| y - z\| $
as claimed.