no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.
for instance, write out pi in base 2, and reinterpret as base 10.
11.0010010000111111011010101000100010000101...
it is infinitely non-repeating, but nowhere will you find a 2.
i've often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven't seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.
The explanation is misdirecting because yes they're removing the penguins from the zoo. But they also interpreted the question as to if the zoo had infinite non-repeating exhibits whether it would NECESSARILY contain penguins. So all they had to show was that the penguins weren't necessary.
By tying the example to pi they seemed to be trying to show something about pi. I don't think that was the intention.
They also say "and reinterpret in base 10". I.e. interpret the base 2 number as a base 10 number (which could theoretically contain 2,3,4,etc). So 10 in that number represents decimal 10 and not binary 10
I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”
They just proved that "X is irrational and non-repeating digits -> can find any sequence in X", as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn't be because of its irrationality or the fact that it's non-repeating, it would be some other property