Likewise, a word or phrase out of Shakespeare, sure. An entire act, not likely. The entire manuscript, no.
The bottom line to me is that Champernowne’s binary constant produces in such a distinctive pattern, it cannot overcome, by 8 bit alphanumeric ASCII concatenation, unreadable characters.
Champernowne's construction (and some that I have invented, too) generate all possible finite bit patterns with the proper frequency. These sequences are not random, but they do satisfy the strong law of large numbers. The frequency of 0 and 1 is 1/2; the frequency of 00,01,10, and 11 is 1/4,; etc.
These sequences can be combined with pseudo-random-number generators to break up obvious correlations in the geneators. It's useful, but still doesn't produce a random sequence.
I think you are looking at it the wrong way.
1) You have a finite string of ASCII bytes of length n
2) View it as just a finite string of bytes of length n -- the fact that it is ASCII encoded is not relevant because the information is preserved.
3) Consider the set of all binary encoded integers of bit-length n*8, where n is the length of the text from (1)
4) The finite set of integers in (3) will, not surprisingly, contain an integer with the exact same bit pattern as the ASCII string in (1).
5) Since Champernownes number is a concatenation of all integers (binary or otherwise -- it doesn't really matter), it will also contain every integer in the set described in (3).
6) Therefore, Champernowne's number will contain every finite string of ASCII bytes, including Shakespeare, and including (1).
Do you see it now? :-)