How to prove a set has infinite cardinality?
A general way of showing this is proof by contradiction.


Assume (towards a contradiction) that the given set $S$ is finite, i.e.

$\qquad S = \{s_1, \dots, s_n\}$.
Show that there is an element $s \in S$ but for which $s \neq s_i$ for all $i \in [1..n]$.

One way to do this is to construct $s$ from $s_1, \dots, s_n$ explicity; that would be an instance of a diagonal argument.
Above step provides a contradiction, hence the assumption must be false: $S$ does not have this form.

Important: This only works if you make no assumptions whatsoever about the $s_i$ and $n$ (beyond what you know about $S$ already). If you do, you have not argued against all finite candidates for $S$.


Example

Show that $\mathbb{N}$ is infinite.


   Assume the opposite. Then, $\mathbb{N} = \{n_1, \dots, n_N\}$ for some $N$. Let $n = \max \mathbb{N}$. Then, by the definition of the natural numbers, $n+1 \in \mathbb{N}$ as well. But that contradicts that $n$ is the largest number in $\mathbb{N}$; the assumption must be false.