What's wrong with this random variable proof? Let $X$ be a Binomial random variable $\sim B(p, n)$. Show that for
  $\lambda &gt; 0$ and $\epsilon &gt; 0$, $P(X - np &gt; n\epsilon) \le
 \mathbb{E}\{\displaystyle e^{\lambda(X - np - n\epsilon)}\}$


I tried like this:

Let $A = \{ \omega \in \Omega: X(\omega) &gt; n(\epsilon + p)\}$.
Then note that 

$$e^{\lambda(X(\omega) - np - n\epsilon)} \ge 1_A(\omega)$$

In fact it is obvious if $\omega \notin A$, and if $\omega \in A$ the right hand side is $1$, and the left hand side is raised to a power $&gt; 0$ so it is itself $&gt; 1$.

So takig the mean both sides yields

$$\mathbb{E}\{\displaystyle e^{\lambda(X - np - n\epsilon)}\} \ge P(A) = P(X - np &gt; n\epsilon)$$ which is what I wanted to prove.

(Because $\mathbb{E}\{1_A\} = P(A)$)

Something has to be wrong though, because I used nowhere the fact that $X$ is a binomial random variable!

What's wrong with it?
You've proven a specific case of what Wikipedia calls the exponential version of Chebyshev's inequality. Note you can pull a $e^{-\lambda n\epsilon}$ out of your expected value on the right hand side; you're equivalently showing:
$$
  P(X - np &gt; n\epsilon) \le e^{-\lambda n\epsilon} E[e^{\lambda(X-np)}],
$$
which is the form given on Wikipedia (and other references, of course) where the RV you're acting on is $Z=X-np$, $\lambda = t$, and your epsilon is $n\epsilon$.  This inequality doesn't require that $X$ be binomial, so there's nothing wrong, even though you didn't use that fact.
Hint:Markov's inequaliy for random variables can be applies to get the result..