Median of Means notes proof wrong? My main question is about the last step in the proof of one property in this short notes about the method of Median of Means(MoM) estimator. The Proposition is stated as follows:

Proposition 1 Assume that $\operatorname{Var}\left(X_{1}\right)<\infty .$ Then the MoM estimator has the following property:
$$
  P\left(\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right) \leq e^{-2 K\left(\frac{1}{2}-\frac{K}{n} \frac{\sigma^{2}}{\varepsilon^{2}}\right)^{2}}=e^{-2 \frac{n}{B}\left(\frac{1}{2}-\frac{\sigma^{2}}{B \varepsilon^{2}}\right)^{2}}
  $$
for every $n=K \cdot B$. Where K is the number of means(number of subsamples), and B is the size of each subsample.

The proof of it proceed as follows:
Since the event $\left\{\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right\}$ implies that at least $K/2$ of $\hat{\mu_l}$ has to be outside $\varepsilon$ distance to $\mu_0$. Namely,
$$
\left\{\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right\} \subset\left\{\sum_{\ell=1}^{K} I\left(\left|\widehat{\mu}_{\ell}-\mu_{0}\right|>\varepsilon\right) \geq \frac{K}{2}\right\}
$$
Define $Z_{\ell}=I\left(\left|\widehat{\mu}_{\ell}-\mu_{0}\right|>\varepsilon\right)$ and let $p_{\varepsilon, B}=\mathbb{E}\left(Z_{\ell}\right)=P\left(\left|\widehat{\mu}_{\ell}-\mu_{0}\right|>\varepsilon\right)$, then the above implies that
$$
\begin{aligned}
P\left(\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right) & \leq P\left(\sum_{\ell=1}^{K} Z_{\ell} \geq \frac{K}{2}\right) \\
&=P\left(\sum_{\ell=1}^{K}\left(Z_{\ell}-\mathbb{E}\left(Z_{\ell}\right)\right) \geq \frac{K}{2}-K p_{\varepsilon, B}\right) \\
&=P\left(\frac{1}{K} \sum_{\ell=1}^{K}\left(Z_{\ell}-\mathbb{E}\left(Z_{\ell}\right)\right) \geq \frac{1}{2}-p_{\varepsilon, B}\right)
\end{aligned}
$$
The key trick of the MoM estimator is that the random variable $Z_{\ell}$ is IID and is bounded. So by Hoeffding's inequality (one-sided),
$$
P\left(\frac{1}{K} \sum_{\ell=1}^{K}\left(Z_{\ell}-\mathbb{E}\left(Z_{\ell}\right)\right) \geq t\right) \leq e^{-2 K t^{2}}
$$
As a result,
$$
\begin{aligned}
P\left(\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right) & \leq P\left(\frac{1}{K} \sum_{\ell=1}^{K}\left(Z_{\ell}-\mathbb{E}\left(Z_{\ell}\right)\right) \geq \frac{1}{2}-p_{\varepsilon, B}\right) \\
& \leq e^{-2 K\left(\frac{1}{2}-p_{\varepsilon, B}\right)^{2}}
\end{aligned}
$$
To conclude that proof, note that the variance $\sigma^{2}=\operatorname{Var}\left(X_{1}\right)<\infty$ and the Chebeshev's inequality implies
$$
p_{\varepsilon, B}=P\left(\left|\widehat{\mu}_{\ell}-\mu_{0}\right|>\varepsilon\right) \leq \frac{\sigma^{2}}{B \varepsilon^{2}}=\frac{K}{n} \frac{\sigma^{2}}{\varepsilon^{2}}\tag{1}
$$
So the bound becomes
$$
P\left(\left|\widehat{\mu}_{\mathrm{MoM}}-\mu_{0}\right|>\varepsilon\right) \leq e^{-2 K\left(\frac{1}{2}-p_{\varepsilon, B}\right)^{2}} \leq e^{-2 K\left(\frac{1}{2}-\frac{K}{n} \frac{\sigma^{2}}{\varepsilon^{2}}\right)^{2}} \tag{2}
$$
My question is at Eq.(2), while the formula before it looks fine to me.  Eq.(1) states $p\le \sigma^2/(B\epsilon^2)$, But there's a $1/2$ factor in the Eq.(2), and if $p_{\epsilon,B}$ is greater than 1/2 at the beginning, then make $p_{\epsilon,B}$ bigger will only cause the term in Eq.(2) to become smaller, which conflicts with the $\le$ notation in the process of replacing of $p_{\epsilon,B}$ into $p\le \sigma^2/(B\epsilon^2)$. So does this proof in the last step, i.e., Eq.(2) wrong? Or it's just that I miss something? Because the references in this short note(chapter 2 of the reference, instead of chapter 3 noted in the note) also did the same proof, so I kind of feel maybe I miss something?

When you use $p_{\epsilon,B}>0.5$ then you are effectively applying Hoeffding's inequality with negative $t$
With Hoeffding's inequality
$$P\left(\frac{1}{K} \sum_{\ell=1}^{K}\left(Z_{\ell}-\mathbb{E}\left(Z_{\ell}\right)\right) \geq t\right) \leq e^{-2 K t^{2}}$$
you do not use $t<0$. For $t = 0$ the upper limiting probability is already $1$.
