Suppose $x$ is a positive real number. It can be shown that $n\mapsto 10^n$ is a map that is unbounded above in the integers, so by Archimedean property, there is some integer $n$ with $10^{n+1}\ge x$. Take the least such $n$ (why must one exist?), and let $a_{-n}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $a\cdot 10^n<x$ (why must one exist?). Now, given $a_{-n},a_{-n+1},...,a_{m}$ for some $m\in\Bbb Z$ with $m\ge-n,$ we let $a_{m+1}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $$a\cdot 10^{-(m+1)}<x-\sum_{j=-m}^na_{-j}\cdot10^j$$ (why must one exist?) Recursively, this determines a sequence $a_{-n},a_{-n+1},...$ of elements of $\{0,1,2,...,8,9\}$ such that for all integers $m\ge-n$ we have $$S_m:=\sum_{j=-m}^na_{-j}\cdot 10^j<x.$$ In fact, $S_{-n},S_{-n+1},...,S_m,...$ is a non-decreasing sequence of positive numbers, so since bounded above by $x,$ this sequence converges to some number no greater than $x$ by the Monotone Convergence Theorem. We can even do better than that, and show that the sequence of partial sums $S_n$ converges to $x$ (why?). The series thus determined is the infinite decimal expansion of $x$.

If $x$ had been negative, we could acquire a decimal expansion for $-x$ in this way, and then the opposite of that would be a decimal expansion for $x$. $0$ has a decimal expansion, too, so all real numbers have a decimal expansion, and moreover, all of them (except arguably $0$) have an infinite decimal expansion (though some also have a finite decimal expansion).