With $s=\sigma+ti$, take the following elegant integral for the Dirichlet $\eta$-function :
$$\eta(s) = \int_{0}^{\infty} \frac{\left(\frac12+ix\right)^{-s}+\left(\frac12-ix\right)^{-s}}{\,\,e^{\pi x}+e^{-\pi x}} dx, \qquad s \in \mathbb{C}$$
and split it into two parts that each remain convergent in $s \in \mathbb{C}$:
$$A1(s) = \int_{0}^{\infty} \frac{\left(\frac12+ix\right)^{-s}}{e^{\pi x}+e^{-\pi x}} dx\qquad A2(s) = \int_{0}^{\infty} \frac{\left(\frac12-ix\right)^{-s}}{e^{\pi x}+e^{-\pi x}} dx$$
Here is a plot of $|A1(1/2+ti)|$ and $|A2(1/2+ti)|$:
that shows $|A1(s)|$ to be an oscillating function and $|A2(s)|$ to be monotonically decreasing for increasing $t$. Zooming in on a non-trivial zero (the 3rd one):
shows that $|A1(s)|$ 'pierces' through $|A2(s)|$ inducing two zeros for $|A1(s)|-|A2(s)|$, of which one, either left or right, is always a non-trivial zero ($\rho$, assuming RH). So, with the ever declining $|A2(s)|$, the function $|A1(s)|$ has to oscillate with an ever increasing amplitude to be able to 'pierce' $|A2(s)|$ and induce an incremental non-trivial zero.
Changing $\sigma$ sightly e.g. $1/2.05$ moves up $|A1(s)|$ a bit, i.e. still maintaining two zeros. Increasing $\sigma$ further will quite quickly make both zeros 'collide' and vanish.
Questions:
Could $|A1(s)|$ ever become zero on its own?
Could it be proven that $|A2(s)|$ is a monotonically decreasing function?