question on equivalent ideas of absolute continuity of measures

Here is a counter-example:

Let $$\nu(E) = \int_E \frac{1}{x^2}dm(x)$$ where $m$ denotes the Lebesgue measure. The measure $\nu$ is not finite.

Then $\nu \ll m$ but $$\exists \epsilon=1>0 \quad \forall \delta>0 \quad \exists E=]0,\delta/2] \text{ measurable set such that } m(E)<\delta$$ but $\nu(A)≥\epsilon$, because $\nu(E) = \left[ \frac{-1}{x} \right]_0^{\delta/2} = \infty$.

One direction of the equivalence always holds, that is, if the $\epsilon$-$\delta$ condition holds, then $\nu<<\mu$, because if $\mu(A)=0$, then for all $\epsilon>0$ and its corresponding $\delta>0$ we have $\mu(A)<\delta\implies \nu(A)<\epsilon$, and hence $\nu(A)=0$.

To see why the finiteness of $\nu$ is important, we therefore must examine the other direction. We can trivially contradict it if we assume $\mu$ to be arbitrarily small, such as the Lebesgue measure, because then we can consistently define a $\mu$-absolutely-continuous measure by $$\nu(A) = \begin{cases}0 & \mu(A)=0\\ \infty & \mu(A)>0\end{cases}$$ However, to gain more insight, we may examine the proof of the other direction when $\nu$ is finite, as presented, e.g., in this answer based on Folland's Real Analysis. The proof hinges on the continuity from above of finite measures, and unsurprisingly, the measure $\nu$ we defined above is a good counterexample to that property.