Cantors diagonal argument tells us that there are uncountably many bisteams. That is to say that a set of all bitstreams cannot be put into a 1 to 1 correspondence with the natural numbers, enumerated for short.
The proof is appears simple. If we could enumerate the bitstreams, then we could generate a list as follows:
\begin{align} \text{Naturals}&&\text{Bitstreams}\\ 0&&\color{red}{0}000\dots\\ 1&&0\color{red}{1}00\dots\\ 2&&10\color{red}{0}0\dots\\ 3&&110\color{red}{0}\dots\\ \end{align}
Take the diagonal of the bitstreams in this list and "flip" each bit. The concatenation of every flipped bit $\color{red}{1011\dots}$ would itself be a bitstream and not in the list. Therefore the assumption that we could enumerate the bitstreams was a false one.
Can we repeat this proof using the bitsreams themselves instead of trying to make a correspondence with the natural numbers?
\begin{equation}\newcommand\mapsfrom{\mathrel{\reflectbox{\ensuremath{\mapsto}}}} \text{bitstreams}\mapsto\text{bitstreams} \end{equation}
That is, can we replace each natural number on the left with the bitstream on the right, and follow the same procedure as before, to show that the bitstreams cannot be put into a 1 to 1 correspondence with themselves?
Elaborating on my comments: I think you haven't asked the question you intended to ask. Of course the set of bitstreams can be put into $1$-to-$1$ correspondence with themselves: you send every bitstream to itself. The question I think you're asking is where the proof of Cantor's theorem fails when you try to repeat it in this situation.
Where it fails is that there is no "diagonal." For there to be a diagonal the number of rows in the table has to be equal to the number of columns. For the enumeration using natural numbers there are countably infinitely many of both. If you try to run the same argument using bitstreams there are still only countably many bits in each bitstream (columns) but uncountably many bitstreams (rows). So you'll find that you can write down a bitstream which is different from any countable collection of bitstreams (this is equivalent to the set of bitstreams being uncountable), but you can't say any more than that.