It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.
The choice $\eta = c_1^{1/10}$ should ensure that the term $\int_{|g-1|\leq \eta} |g|^\beta \operatorname{sgn}(g) \, d\mu$ converges to 1 for $c_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.
To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align}
\mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c_1)) \leq \mu(|g-\mathbb{E}[g]| > c_1^{1/3}) \rightarrow 0
\end{align}
for $c_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.
