Alright, so, the other proof that I promised:

If we define 0.999... as the sum of the series 9/10+9/100+9/1000+..., then for every neighbourhood U(1) it is true that there exists a metric ball B_N = B(1, 1/10^N), where N is natural, such that B_N is a subset of U(1).

For all natural n > N it is true that d(sum(9/10^k) for k from 1 to n, 1) = |1 - sum(9/10^k) for k from 1 to n| = |1/10^n| = 1/10^n < 1/10^N, meaning that for all natural n > N it is true that sum(9/10^k) for k from 1 to n is in B_N, meaning that it is also in U(1).

However, sum(9/10^k) for k from 1 to n is the nth partial sum of the series 9/10+9/100+9/1000+..., which, together with the fact that every such sum is in U(1) for n > N, means that 1 is the limit of the sequence of the partial sums of the series 9/10+9/100+9/1000+..., meaning that 1 is the sum of that series. That means that 0.999... is 1 by definition.