Recall that $\lim_{x\to a}\ f(x)=L$ means that for all $\epsilon>0$ we are able to find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$.

In the case of $\lim_{x\to a}\ x=a$ this means that for all $\epsilon>0$ we have to find a $\delta>0$ such that
$\ \ \ \ \ \ \ (*)\ \ \ $ if $0<|x-a|<\delta$ then $|x-a|<\epsilon$.

So let $\epsilon$ be any positive real number.

To make $(*)$ true, we can choose $\delta=\epsilon$.

With this (or any smaller) choice of $\delta$, it is true that if $|x-a|<\delta$ then $|x-a|<\epsilon$.

This proves that $\lim_{x\to a}\ x=a$.