Continued fractions and irrationality

Setup. Write $\sqrt2=1+(\sqrt2-1)$, so
$$\sqrt2=1+\frac{1}{\sqrt2+1}.$$
And $\sqrt2+1=2+(\sqrt2-1)$, so
$$\sqrt2=1+\cfrac{1}{2+\cfrac{1}{\sqrt2+1}}=1+\cfrac{1}{2+\cfrac{1}{2+\cdots}}=[1;\overline{2}].$$

Irrationality. Rationals have finite continued fractions. $\sqrt2$'s expansion is infinite and periodic, so $\sqrt2$ is irrational.

Conclusion. $\sqrt2=[1;\overline2]$ is irrational.