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).
This is a good start, but if I understand your method correctly, it does not quite work. It proves that you can represent infinitely many numbers— but it does not clearly show that you can represent all numbers.
For contrast, consider a similar claim:
You can represent any integer using a binary number whose unit bit is 0.
This claim is false because if the unit bit is 0, you can only represent even numbers. However, according to your proposed proof strategy, you might argue: (a) in the base case, you have one bit and can represent one number. (b) if you have $(k+1)$ bits, you can simply add $2^k$ to each of the numbers you have represented so far to get twice as many new numbers. This is true, but it only implies that you can represent infinitely many numbers in this scheme— in fact, you can represent all the even numbers. It does not imply that you can represent all integers, however. And indeed you can't represent odd numbers this way.
If you like, you could treat your goal formally as establishing a bijection between the natural numbers $\mathbb{N}$ and binary strings $\mathbb{B}$.
Thus, you would need to show that every integer has a bitstring representation, and that bitstring representations are unique— two different bitstrings must correspond to different numbers. (I'm assuming that leading zeroes are forbidden.)
But perhaps you only want to prove that every integer has a corresponding bitstring representation, and you don't care about uniqueness.
This is easy enough to show— we just convert each number to binary! Let $n$ be a non-negative integer.
Put $n_0 = n$ and $b_0 = n\pmod{2}$. Until $n_k$ is zero, define $n_{k+1} = (n_k - b_k)/2$ and define $b_{k+1} = n_{k+1}\pmod{2}$.
Because the $n_k$ form a decreasing sequence of non-negative numbers, this process must eventually end. When it does, declare the bitstring of $n$ to be $b_k\ldots b_0$. This is the binary representation of $n$. Because you can apply this procedure to any number $n$, it proves that every number has a binary representation.
This proof does not show a few other things, namely that bitstrings are unique (each number has at most one representation), or that every bitstring corresponds to a valid number.
Best Answer
Every real number $\alpha>0$ (to make things simpler) can be developed into an infinite binary expansion of the form $$110\ldots10\,.\,1011101\ldots\ ,$$ where to the left of the "decimal" point we only have finitely many digits and to the right an unending sequence of digits. Numbers of the form $n/2^N$ $\ (n, N\in{\mathbb N}_{\geq0})$ have exactly two such representations (this is the $0.999\ldots=1.000\ldots$ phenomenon in decimal), all other numbers have exactly one representation. In fact it is possible to construct the full real number system (with addition, multiplication and order) as such an "uncountable list" of binary fractions. This has been sketched by Gowers here: http://www.dpmms.cam.ac.uk/~wtg10/decimals.html
A real number $\alpha>0$ can be considered as "given" when for each $N$ it is possible to name an integer $a_N$ such that $${a_N\over 2^N}\leq \alpha<{a_N+1\over 2^N}\ .$$ Using induction one easily proves that $$2 a_N\leq a_{N+1}\leq 2a_N+1\ ,$$ and this implies that the finite binary representation of $a_{N+1}$ is obtained from the representation of $a_N$ by appending a $0$ or a $1$. Now the quotients $a_N/2^N$ approximate the given number $\alpha$. Writing $a_N$ in binary and separating the last $N$ digits by a "decimal" point we therefore get a finite binary approximation of $\alpha$, and things work out such that for $N'>N$ the first $N$ places (after the "decimal" point) of $a_N$ and $a_{N'}$ coincide. In this way we get an ever better approximation of $\alpha$ just by adding new "binary places" to the right.