Rigorous proof that $\frac{1}{3} = 0.333\ldots$

Since $\sum_{i=1}^\infty 3\cdot 10^{-i}$ is what the notation "$0.333...$" means, your argument is perfectly good. It's not just the "start of a proof", it is all there is to it.

Okay, perhaps it is not really trivial to prove that the geometric series converges, but straightforward it is. Just plug in the definition of the sum of a series and crank the handle, using standard tricks to rewrite each of the partial sums in turn.

Your proof is correct. Essentially you proved it was the convergence of the geometric series.

Alternatively if you want to avoid using the geometric series formula and just use a very simple limit argument, note that , $(3)(.333...) = .999... = 1$ $\text{ }$ To see this, note that $.999... = \sum_{i = 1}^\infty \frac{9}{10^i}$ converges to $1$. Take a look at partial sums. Thus $.333... = \frac{1}{3}$.