How to turn adversary against encryption of zeroes into adversary against random bits?
It's actually quite simple.
Given your adversary $A$ against ind you construct an adversary $A'$ against ind$ by simply forwarding the queries and answers. (This may seem stupid but please bear with me.)

Consider now, the difference between the advantage of $A$ and $A'$:
$$\mathrm{Pr}[K \xleftarrow{$} \mathrm{Key} : A ^{\mathcal{E}_K(\cdot)} \Rightarrow 1] - \mathrm{Pr}[A ^{\mathcal{$}(\cdot)} \Rightarrow 1]-(\mathrm{Pr}[K \xleftarrow{$} \mathrm{Key} : A ^{\mathcal{E}_K(\cdot)} \Rightarrow 1] - \mathrm{Pr}[K \xleftarrow{$} \mathrm{Key} : A ^{\mathcal{E}_K(0^{|\cdot|})} \Rightarrow 1])$$
$$= \mathrm{Pr}[K \xleftarrow{$} \mathrm{Key} : A ^{\mathcal{E}_K(0^{|\cdot|})} \Rightarrow 1] - \mathrm{Pr}[A ^{\mathcal{$}(\cdot)} \Rightarrow 1]$$

If this distance were non-negligible, we could build a successful adversary $A''$ against ind$\mathrm{$}$ as follows: Whenever $A$ queries a message, $A''$ queries $0^{|\cdot|}$ instead and returns the answer to $A$.

It's easy to see that the advantage of $A''$ is excactly the difference from above. Therefore you get $$\mathbf{Adv}_{\Pi}^{\mathrm{ind$}} = \mathbf{Adv}_{\Pi}^{\mathrm{ind}}  - \mathbf{Adv}_{\Pi}^{\mathrm{ind$}}$$
$$\Rightarrow 2\cdot\mathbf{Adv}_{\Pi}^{\mathrm{ind$}} = \mathbf{Adv}_{\Pi}^{\mathrm{ind}}$$
$$\Rightarrow\mathbf{Adv}_{\Pi}^{\mathrm{ind$}} = \frac{1}{2}\cdot\mathbf{Adv}_{\Pi}^{\mathrm{ind}}$$