Proving $S:=\left\{1\Big/\left(x^2-3\right)\mid x\in\Bbb Q\right\}$ is unbounded 
Let $S:=\left\{\dfrac1{x^2 -3} :x\in\Bbb Q\right\}$. Prove
carefully that $S$ is unbounded

My proof
By contradiction, if we can find some $M$ so that $|s| \leqslant M\ \ \forall s \in S$, that is, if
$$ \dfrac{1}{|x^2-3|} \leqslant M\ \forall x\in\Bbb Q\ . $$ Since $x$ is any rational number, we can choose $x$ so that $|x^2-3| = \dfrac{1}{M+1}$. Therefore, we obtain
$$ \dfrac{1}{\frac{1}{M}+1} \leqslant M \implies M+1 \leqslant M $$
A contradiction.
I think my proof is correct, but I have lost some points. What is my mistake?

One could show that $S$ is neither bounded from above nor bounded from below. However, we are only asked to sow that it is not bounded. This can be done ab ovo without any sophisticated converging sequences a la Heron formula.
So assume there exists $M\in \Bbb R$ with $|s|\le M$ for all $s\in S$.
Pick $n\in\Bbb N$ with $n>2M$. Then $A_n:=\{\,x\in\Bbb N\mid x^2>3n^2\,\}$ is a non-empty (e.g., $2n\in A_n$) proper (e.g., $1\notin A_n$) subset of $\Bbb N$ and we can let $m=\min A_n$. Then $$m^2>3n^2$$ and $(m-1)^2\le 3n^2$. We readily see (as in the proof of irrationality of $3$) that even stronger $$(m-1)^2<3n^2.$$
From $(2n)^2>3n^2$, it is also clear that $$ m\le 2n.$$
With $x_1=\frac{m-1}n$ and $x_2=\frac mn$, we have
$$\begin{align}\frac1{x_2^2-3}-\frac1{x_1^2-3}&=\frac{n^2}{m^2-3n^2}-\frac{n^2}{\underbrace{(m-1)^2-3n^2}_{<0}}\\
&=\frac{n^2}{m^2-3n^2}+\frac{n^2}{3n^2-(m-1)^2}  \\
&=\frac{n^2(3n^2-(m-1)^2+m^2-3n^2)}{(m^2-3n^3)(3n^2-(m-1)^2)}\\
&=\frac{(2m-1)n^2}{(m^2-3n^3)(3n^2-(m-1)^2)}.\end{align}$$
By the arithmetic-geometric inequality, the denominator is $$ \begin{align}(m^2-3n^3)(3n^2-(m-1)^2)&\le\left(\frac{m^2-3n^3+3n^2-(m-1)^2}{2}\right)^2\\&=\frac{(2m-1)^2}{4}\end{align}$$
so that 
$$2M\ge \frac1{x_2^2-3}-\frac1{x_1^2-3}\ge\frac{4n^2}{2m-1}
\ge \frac{4n^2}{4n-1}>n.$$
Contradiction.
