I thought this proof was much simpler than it actually is. I used an Axiom that states, "if $p$ and $q$ are real numbers, then there is a number between them, i.e: $$\frac{(p + q)}{2}$$ My attempt at the proof was:
By the above Axiom, there exists a number, $q$, between $x$ and $y$, i.e: $$\frac{(x+y)}{2}$$ We know $q$ is rational because it is in the form $\frac ab$, where $a = x + y$ and $b = 2$. Thus, there exists a rational number, $q$ such that $q$ is between $x$ and $y$.
However, I realized that if $x=\pi$ and $y = e$, then $\frac{(x+y)}{2}$ will not be rational.
I know that to prove this, I need to use the Lemma: If $y-x>1$, ∃ n ∈ $\mathbb{Z}$ such that $x<n<y$.
I really don't know how to approach this differently because I did think it would be simpler.