I got a proof of this but I am quite sure that it is not what was expected on the exam. Also, this proof seems really kludgy and non-kosher.
Because of the density of the raitonals in the reals, there exists a $q\in\mathbb{Q}$ such that $\sqrt3-\frac1 n < q < \sqrt3$
For each n, let $a_n = q$. ($\sqrt3-\frac1 n < a_n < \sqrt3$)
So for n=1. There exists a $q$ such that $\sqrt3-1 < q < \sqrt3$. Choose this q for $a_1$
For n=2, choose $q$ such that $\sqrt3-\frac1 2 < q < \sqrt3$.
Since $\sqrt3-\frac1 n$ converges to $\sqrt3$, $a_n$ must also converge to $\sqrt3$.
So we are constructing a sequence out of things that we are only know the existence of. Also does this require the axiom of choice?
Anyways, on the exam there was a hint "consider $S = \{r\in\mathbb{Q}|r>0 \,\mathrm{and}\,r^2<3\}$" And I am not exctly sure what to make of it other than the fact that $\sup S = \sqrt 3$
Edit: To be specific about my question, is this proof ok? And what is the standard proof (the proof that my prof was hinting at)?