
This means there is no sequence x_{1}, x_{2}, x_{3},... that contains all the real numbers.
Suppose x_{1}, x_{2}, x_{3},... has all of R as its range.
Start by finding the first two elements of x that are different from one another. Call the first one x_{i} and the second x_{j}. Let a_{1}=min(x_{i},x_{j}) and b_{1}=max(x_{i},x_{j}). Consider the first "j" elements of sequence x as having been "examined".
Now, having found a_{n} and b_{n}, examine elements of sequence x that haven't yet been examined until you find two, x_{i} and x_{j}, that are different from one another and also between a_{n} and b_{n}. Let a_{n+1}=min(x_{i},x_{j}) and b_{n+1}=max(x_{i},x_{j}).
The two strictly monotone sequences, a and b, move toward each other.
By completeness of the reals, since sequence a is bounded above, sequence a has a "supremum", or least upper bound, c. c is larger than every element of a, because if c=a_{n} for any n, then c < a_{n+1} and so c would not be the supremum of sequence a.
c is smaller than every element of sequence b, because if c≥b_{n} for any n, then b_{n+1} (which is smaller than b_{n}) would be a smaller upper bound than c of sequence a, and so c would not be the least upper bound of sequence a.
Since c is larger than every element of a and smaller than every element of b, it follows that c is contained in every open interval (a_{n},b_{n}). If c were equal to any element of sequence x, say, x_{i}, then upon examination of x_{i}, c=x_{i} would have been one of the two numbers chosen as a_{n+1} or b_{n+1}, contradicting the inclusion of c in every open interval.
Therefore, c is not one of x_{1}, x_{2}, x_{3},..., contradicting the assertion that sequence x has all of R as its range.
Completeness, from the Wikipedia
Cantor's first uncountability proof, from the Wikipedia
Upper Bound and Least Upper Bound (Supremum)
The webmaster and author of this Math Help site is Graeme McRae.