This claim is proven by building a bijection
by following a diagonal procedure.
Imagine you have an infinite matrix, whose coefficient are all possible fractions:
![\left[\begin{array}{cccc}(1)/(1)&(2)/(1)&(3)/(1)&\ldots\\&&\\(1)/(2)&(2)/(2)&(3)/(2)&\ldots\\&&\\(1)/(3)&(2)/(3)&(3)/(3)&\ldots\end{array}\right]](https://img.qammunity.org/2019/formulas/mathematics/high-school/jbmdsriztcaumr70fd0vjghya3yvdm90ey.png)
Now, count them "diagonally": this is how we map every natural number, using the same positions:
![\left[\begin{array}{cccc}\phi(1)&\phi(2)&\phi(6)&\phi(7)\\\phi(3)&\phi(5)&\phi(8)&\phi(13)\\\phi(4)&\phi(9)&\phi(12)&\ldots\\\phi(10)&\phi(11)&\ldots&\ldots\end{array}\right]](https://img.qammunity.org/2019/formulas/mathematics/high-school/3lu99rkaoe8djqtuz79r8080r669q9ekcg.png)
The attached image, taken from aminsaied.wordpress.com/2012/05/21/diagonal-arguments/, should make it clearer