Yeah I have no idea what the author is getting at there. One obvious way to put a topology on the space of binary streams is to think of them as binary expansions of real numbers in the interval [0, 1] (the notation is supposed to indicate that we include 0 and 1) and inherit the topology from the normal topology on the reals in which case the answer is yes [0,1] is compact.
In this representation 0 is included as the binary stream consisting of all zeros and 1 is included as the binary stream consisting of all 1s.
This isn't entirely clean as a bunch of (rational) real numbers in [0,1] have multiple different binary expansions, for example 1/2 can be written
In particular, you can search compact infinite spaces by only considering a finite number of steps: "What is going on here is that computable functionals are continuous, which amounts to saying that finite amounts of the output depend only on finite amounts of the input. But the Cantor space is compact, and in analysis and topology there is a theorem that says that continuous functions defined on a compact space are uniformly continuous. In this context, this amounts to the existence of a single `n` such that for all inputs it is enough to look at depth `n` to get the answer (which in this case is always finite, because it is an integer). I’ll explain all this in another post. Here I will illustrate this by running the program in some examples."
You can avoid the problem of multiple expansions by using base 3 numbers that don't contain the digit 2. The resulting space is equivalent to the Cantor space mentioned in other comments.
In this representation 0 is included as the binary stream consisting of all zeros and 1 is included as the binary stream consisting of all 1s.
This isn't entirely clean as a bunch of (rational) real numbers in [0,1] have multiple different binary expansions, for example 1/2 can be written
.10000... (the zeros go on forever)
or
.01111... (the ones go on forever)