Demonstrate that there are no perfect squares ending with $8$ A number n will always end in some digit of the set {$0,1,2,3,4,5,6,7,8,9$}. The last digit of $n^2$ is the last digit of its last squared digit. Like this:
$$\ldots 0^2 = \ldots 0$$
$$\ldots 1^2 = \ldots 1$$
$$\ldots 2^2 = \ldots 4$$
$$\ldots 3^2 = \ldots 9$$
$$\ldots 4^2 = \ldots 6$$
$$\ldots 5^2 = \ldots 5$$
$$\ldots 6^2 = \ldots 6$$
$$\ldots 7^2 = \ldots 9$$
$$\ldots 8^2 = \ldots 4$$
$$\ldots 9^2 = \ldots 1$$
Therefore, no perfect square ends in $8.$
I think my proof is pretty bad, is there anything more formal than that?

Your proof is not bad.  You could make a more formal argument with modular arithmetic. 
For example,  $ n\equiv 0, \pm1, $ or $\pm2 \pmod 5$, 
so $n^2\equiv 0, 1, $ or $-1\pmod 5$, 
so $5\nmid n^2-3,$ so $10\nmid n^2-8$.
