Would subsequences of digits of $\pi$ be normal if $\pi$ were itself normal?
Suppose the digits of $\pi$ in base $2$ were normal and were broken down into two subsequences. Would these necessarily be normal?
i.e. Let $\displaystyle\pi=\sum_{n=-1}^{\infty}2^{-2n} a_n+\sum_{n=1}^{\infty}2^{1-2n}b_n$
$a_n\in\{0,1\},b_n\in\{0,1\}\forall n$
If $\pi$ is normal does this imply $a_{-1},a_{0},\ldots$ and $b_{-1},b_{0},\ldots$ are normal?
The converse clearly does not hold, at least in general, because we could have a pair of normal sequences $a_n=b_n$ and the resulting sum would not be normal in base $2$
Yes, in your case this can be done. For instance, if we wanted to show that $11101$ appears somewhere in the sequence $a_i$, then we could just look through the expansion of $\pi$ for any sequence of the form $$1\_1\_1\_0\_11\_1\_1\_0\_1$$ where the underscores are wildcards (note that this is the sequence we want, repeated twice, with blanks put in between each bit except between the repeats themselves). Then, since every other term is in $a_i$, and every other term is in $b_i$, no matter where this sequence starts we know that the first instance of $11101$ is in either $a_i$ or $b_i$, while the other instance is in the other sequence. Thus we get that $b_i$ necessarily contains $11101$ as well.
This trick works as long as the successive differences between the indices of each of the subsequences follow some (eventually) repeating pattern. In this case, the pattern is $2, 2, 2, \ldots$, since each successive pair of elements $a_i, a_{i+1}$ are two indices away from one another in the original sequence, and the same for $b_i$. If the pattern gets more complicated, then so does the new sequence you have to look for.
Note that I could've interleaved the two sequences above and just gone with $1111110011$, but then it would perhaps not have been so easy to see how to use this for less regular patterns.
As a second example, consider if we let $b_i$ have every element whose index is divisible by $3$, and let $a_i$ contain the rest. Then to show that $a_i$ contains the sequence, you look for something like $$ 11\_10\_1\_11\_01\_\ \_\ \_11\_10\_1 $$ in the expansion of $\pi$. Now, no matter how this aligns with the $aabaab$-pattern of which bit goes into which subsequence, somewhere along that sequence, the $a_i$ will pick up $11101$.
Here the successive differences for the $a_i$ are $1, 2, 1, 2,\ldots$, and for $b_i$ it's $3, 3, 3,\ldots$.
As for irregular patterns of differences, if the differences for $a_i$ are unbounded (say $a_i$ contains all the prime index bits, or even something as simple as letting $a_i$ have the triangle number bits, and $b_i$ the rest), then it is clearly not true in general, as you can just make sure that all the $a_i$ are $0$. I honestly don't know whether this is true if the differences are irregular but bounded.