Prove: $a_n \leqslant x$ $\forall_{n \geqslant 1} \Rightarrow \lim_{n \to \infty}$ $a_n \leqslant x$

I'm not sure about how to do the proof of this exercise of my math study. Its exercise 5.4.8 of Analysis I by Terence Tao:

Let $(a_n) _{n=1}^{\infty}$ be a Cauchy sequence of rationals and $x \in \mathbb{R}$

Prove: $a_n \leqslant x$$\forall_{n \geqslant 1} \Rightarrow \lim_{n \to \infty}$$a_n \leqslant x$

As a hint, they tell me to use a contradiction and the theorem that $\forall$$x < y \in \mathbb{R}$, $\exists$$q \in \mathbb{Q}$ for which $x < q < y$

This is what I've done so far:

Let $a_n \leqslant x$$\forall_{n \geqslant 1}$, and assume $\lim_{n \to \infty}$$a_n \nleqslant x$
$\Rightarrow \lim_{n \to \infty}$$a_n > x$
$\Rightarrow \exists$$q \in \mathbb{Q}$ with $x < q < \lim_{n \to \infty}$$a_n$

Can you explain me how to complete the prove?

Thanks in advance!