<s>[INST] What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:

Theorem:  $\sqrt{2}$ is irrational.
Proof:
  $3^2-2\cdot 2^2 = 1$.

(That's it)
That is a corollary of
this result:

Theorem:
  If $n$ is a positive integer
  and there are positive integers
  $x$ and $y$ such that
  $x^2-ny^2 = 1$,
  then
  $\sqrt{n}$ is irrational.

The proof is in two parts,
each of which
has a one line proof.
Part 1:

Lemma: If 
  $x^2-ny^2 = 1$,
  then there are arbitrarily large integers
  $u$ and $v$ such that
  $u^2-nv^2 = 1$.

Proof of part 1:

Apply the identity
  $(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
  as many times as needed.

Part 2:

Lemma:  If
  $x^2-ny^2 = 1$
  and
  $\sqrt{n} = \frac{a}{b}$
  then
  $x < b$.

Proof of part 2:

$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
  or
  $b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
  so
  $x < b$.

These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
 is that
if $n > 0$
then
$n \ge 1$.
 [/INST] Below is a simple way to implement the (Euclidean) denominator descent implicit in Ivo Terek's answer (which was John Conway's favorite way to present this proof).
Theorem $\quad \rm r = \sqrt{n}\:$ is an $\rm\color{#c00}{integer}$ if rational,$\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \ \ $  Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0.\,$ $\  \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ $ Taking fractional parts:  $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ for $\rm\ 0 \le b < B.\ $ If  $\,\rm\displaystyle\ \color{#c00}{B\nmid A}\,$ then $\rm\ b\ne 0,\ $ so $\rm\,\ \displaystyle \frac{A}B = \frac{a}b,\ $ contra leastness of $\,\rm B$.
Remark $\ $ See here for a conceptual view of the proof (principality of denominator ideals). Though the proof may seem unusual to those who have not yet studied advanced number theory, it is quite natural once one learns about conductor and denominator ideals.
</s>