[proofplan]
The error count $S_n$ is the sum of $n$ i.i.d. Bernoulli$(p)$ indicators, so $S_n/n$ is the sample mean of the $(U_i)$. The event $\{S_n \geq n(p + \varepsilon)\}$ is exactly the event $\{S_n/n - p \geq \varepsilon\}$, a one-sided large deviation of the sample mean from its mean $p$. By the [Weak Law of Large Numbers](/theorems/???), $S_n/n \to p$ in probability, which forces the probability of this event to zero.
[/proofplan]
[step:Introduce the error indicators and identify $S_n/n$ as a sample mean]
For each $i \geq 1$, let $U_i: \Omega \to \{0, 1\}$ be the indicator that the $i$-th transmitted bit is received incorrectly. By the definition of a binary symmetric channel with crossover probability $p$, each $U_i$ is Bernoulli$(p)$, and the channel's memorylessness translates into independence of the $U_i$. Thus $(U_i)_{i \geq 1}$ is an i.i.d. Bernoulli$(p)$ sequence, with
\begin{align*}
\mathbb{E}[U_i] = p, \qquad \operatorname{Var}(U_i) = p(1 - p).
\end{align*}
The number of errors in the first $n$ transmissions is $S_n = \sum_{i=1}^n U_i$, so the event of interest can be rewritten as
\begin{align*}
\{ S_n \geq n(p + \varepsilon) \} = \left\{ \frac{1}{n}\sum_{i=1}^n U_i - p \geq \varepsilon \right\}.
\end{align*}
[/step]
[step:Control the two-sided deviation by Chebyshev's inequality]
The random variable $S_n/n$ has mean $p$ and, by independence and [Bienaymé's Identity](/theorems/???), variance
\begin{align*}
\operatorname{Var}\!\left(\frac{S_n}{n}\right) = \frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}(U_i) = \frac{p(1-p)}{n}.
\end{align*}
Since $U_1 \in L^2$ (indeed it is bounded), $S_n/n \in L^2$ and [Chebyshev's Inequality](/theorems/???) applies: for every $\varepsilon > 0$,
\begin{align*}
\mathbb{P}\!\left( \left|\frac{S_n}{n} - p\right| \geq \varepsilon \right) \leq \frac{\operatorname{Var}(S_n/n)}{\varepsilon^2} = \frac{p(1-p)}{n\varepsilon^2}.
\end{align*}
[guided]
We wish to bound the one-sided deviation probability $\mathbb{P}(S_n/n - p \geq \varepsilon)$, but Chebyshev's inequality provides only a two-sided bound in terms of $|S_n/n - p|$. This is not a problem: the one-sided event is contained in the two-sided one,
\begin{align*}
\left\{ \frac{S_n}{n} - p \geq \varepsilon \right\} \subseteq \left\{ \left|\frac{S_n}{n} - p\right| \geq \varepsilon \right\},
\end{align*}
so monotonicity of $\mathbb{P}$ gives a bound on the smaller event from a bound on the larger.
To apply Chebyshev we need the second moment. $U_i$ is bounded by $1$, so $U_i \in L^2$ with
\begin{align*}
\operatorname{Var}(U_i) = \mathbb{E}[U_i^2] - \mathbb{E}[U_i]^2 = p - p^2 = p(1-p),
\end{align*}
using $U_i^2 = U_i$ since $U_i \in \{0, 1\}$. Since the $U_i$ are independent they are pairwise uncorrelated, and Bienaymé's Identity gives
\begin{align*}
\operatorname{Var}\!\left(\frac{S_n}{n}\right) = \frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}(U_i) = \frac{p(1-p)}{n}.
\end{align*}
Chebyshev's inequality then yields
\begin{align*}
\mathbb{P}\!\left( \left|\frac{S_n}{n} - p\right| \geq \varepsilon \right) \leq \frac{p(1-p)}{n \varepsilon^2}.
\end{align*}
[/guided]
[/step]
[step:Pass to the limit and conclude]
Combining the inclusion
\begin{align*}
\left\{ \frac{S_n}{n} - p \geq \varepsilon \right\} \subseteq \left\{ \left|\frac{S_n}{n} - p\right| \geq \varepsilon \right\}
\end{align*}
with the Chebyshev bound from the previous step, we obtain
\begin{align*}
0 \leq \mathbb{P}\!\left( S_n \geq n(p + \varepsilon) \right) = \mathbb{P}\!\left( \frac{S_n}{n} - p \geq \varepsilon \right) \leq \frac{p(1-p)}{n\varepsilon^2}.
\end{align*}
The right-hand side tends to $0$ as $n \to \infty$ since $p(1-p)/\varepsilon^2$ is a fixed finite constant. By the squeeze theorem for real sequences,
\begin{align*}
\lim_{n \to \infty} \mathbb{P}\!\left( S_n \geq n(p + \varepsilon) \right) = 0,
\end{align*}
as required. (Equivalently, this is the conclusion of the [Weak Law of Large Numbers](/theorems/???) applied to the i.i.d. $L^2$ sequence $(U_i)$, specialised to a one-sided deviation.)
[/step]
