I came across this meme today:

The counterproof is very trivial, but I see no one disproves it. Some even say that the meme might be true. Well, $\pi$ cannot contain itself.

Well, everything means $\pi$ might contain $\pi$ somewhere in it. Say it starts going $\pi=3.1415...31415...$ again on the $p$ digit. Then it will have to do the same at the $2p$ digit, since the "nested $\pi$" also contains another $\pi$ in it. $\pi$ then will be rational, which is wrong. Thus $\pi$ does not contain all possible combination.

Is this proof correct? I'm not a mathematician so I'm afraid I make silly mistakes.