This means there is no sequence x1, x2, x3,... that contains all the real numbers.
Suppose x1, x2, x3,... 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 xi and the second xj. Let a1=min(xi,xj) and b1=max(xi,xj). Consider the first "j" elements of sequence x as having been "examined".
Now, having found an and bn, examine elements of sequence x that haven't yet been examined until you find two, xi and xj, that are different from one another and also between an and bn. Let an+1=min(xi,xj) and bn+1=max(xi,xj).
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=an for any n, then c < an+1 and so c would not be the supremum of sequence a.
c is smaller than every element of sequence b, because if c≥bn for any n, then bn+1 (which is smaller than bn) 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 (an,bn). If c were equal to any element of sequence x, say, xi, then upon examination of xi, c=xi would have been one of the two numbers chosen as an+1 or bn+1, contradicting the inclusion of c in every open interval.
Therefore, c is not one of x1, x2, x3,..., 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)
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