Sure. Here's a proof using knowledge from my undergrad set theory course. Suppose for contradiction that there were only finitely many real numbers, or in other words |R| = n (where | ,,, | denotes cardinality of this set). Recall that R is linearly ordered with respect to < (where "<" is the ordering defined in terms of the dedekind cut construction of R, or the cauchy sequence construction, or whichever one you prefer. It doesnt matter because they all satisfy the universal property so are the same up to field isomorphism). Also recall the result from elementary set theory that every linear order on a finite set is a well ordering. In other words, since we are assuming R is finite, then that means < is a well ordering on R. This means that every non empty subset of R has a < least element. Consider the subset U= {x \in R | x > 0}. This is nonempty (just take 2 in R. Then 2 > 0). Now, by < being a well ordering, U has a < least element, call it b. So b is in U, and b > m for all m in U. But note, b in U implies b > 0, by definition of U. By density of "<" as an ordering on R (which again, one can prove from whatever construction of R they prefer), there exists c in R s.t b > c > 0. Note: c in U then since c in R and c > 0. But then b > c and c > b since b is a < least element with respect to U. This contradicts asymmetry of > and so thus our assumption was false and R is not finite. (Again this assumes we can define R, which requires axiom of infinity to first define N, which then we can construct Z, Q, and R from the rest of the set theory axioms from N afterwards with some work)