Show probability bound. Let $Y$ be the number of successes in $n$ trials of a Bernoulli experiment with success probability $p$.  Show that:
$$ Pr(|\frac{Y}{n} - p |<e) \geq 1 - \frac{1}{4ne^2}$$
I tried starting with Chebychev and work backwards but I got stuck:
I used $np(1-p)$ for variance since binomial.
$$
\begin{align}
Pr(|Y-\mu|\geq a) &\leq \frac{np(1-p)}{a^2}\tag1\\
Pr(|\frac{Y}{n}-\frac{\mu}{n}|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag2\\
\end{align}
$$
Since for binomial $\mu/n=p$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag3\\
1-Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag4\\
Pr(|\frac{Y}{n}-p|< \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag5\\
\end{align}
$$
I then set $e=a/n$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{np(1-p)}{(en)^2}\tag6\\
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{p(1-p)}{e^2n}\tag7\\
Pr(|\frac{Y}{n}-p|< e) &\geq \frac{e^2n-p(1-p)}{e^2n}\tag8\\
\end{align}
$$
I gave up here because I still need to flip the greater than in the prob and get the RHS to be exactly the same (how??) and I still don't really know if this is the right approach to start with anyway.

Check out Hoeffding's inequality on Wikipedia. It won't give you exactly the answer that you're looking for: instead, it will give you a far stronger one (asymptotically, as $n$ gets large)! Regarding the precise formlation that you have, you're almost there: all you need to do is note that
$$ p(1-p) \le \tfrac14 \quad \text{for all} \quad p \in [0,1]. $$
I'd definitely advise looking at the Wiki page though. There you'll find a proof of a slightly more general result -- it generalised from Binomial, ie sum of independent Bernoullis, to a sum of independent bounded random variables.
I leave the proof to your reading of that page, and just look here at how to apply it. It says that
$$ P( |Bin(n,p) - np| \ge \epsilon n ) \le 2 \exp(-2 \epsilon^2 n). $$
Note that, $|Y/n - p| < e$ if and only if $|Y - pn| < en$. Hence, applying this, we find that
$$ P( |Y/n - p| \ge e ) = P( |Y - pn| \ge en ) \le 2 \exp(- 2 e^2 n). $$
Taking the compliment gives the probability you desire.
This actually shows a much better bound that you've been asked to show for large $n$. Note that Hoeffding is valid for all $n$.
