Let r1 r2 r3 … be a Cauchy sequence of real numbers, i.e. a Cauchy sequence of Cauchy sequences of rationals. Think of r as a matrix, where ri,j is the jth term in the ith sequence. Build a new sequence s as follows.
For a given n, find the tail of rn, such that every pair of terms contained in that tail has a separation less than 1/n. If c is the first term of this tail, let sn = c.
We need to show s is Cauchy. Given an ε, find k such that the limit of ri and the limit of rj differ by less than ε whenever i and j are greater than k. Then find n such that 1/n is less than ε. If k is less than n, advance k to n. Now assume k ≤ i ≤ j. We know that the tail of ri, and hence the limit of ri, is trapped within si±1/n, or si±ε. The limit of ri and the limit of rj are within ε of each other. And the limit of rj is within ε of sj. Put these three together and si-sj is no more than 3ε, for all i and j beyond k. Let ε shrink to 0, and s is Cauchy. Hence s represents a real number.
Is s the limit of our sequence of real numbers? Given ε, compute k as above. Then increase k as necessary, so that the terms of s, beyond k, are within ε of each other. This can be done since s is Cauchy. Now let j be any index beyond k. Let e be the difference sequence s-rj. For i beyond k, look at the term si-rj,i. We know si and sj are within ε of each other, and as long as we stay within the tail of rj, sj and rj,i are within ε of each other. Differences are bounded by 2ε, so the limit of s-rj is bounded by 2ε. This holds for all j beyond k, and we can do this for any ε. Therefore the real number s is the limit of our sequence of real numbers, and the reals are complete.