How to construct a next bit predictor from a distinguisher
Your idea for constructing a distinguisher from a predictor is fine, assuming you know that the predictor predicts the last bit.  The more general statement is: if you can predict any bit of the output, say the $i$th bit, given the first $i-1$ bits, then you can also build a distinguisher.  A similar idea to what you showed also works to prove this statement.

For constructing a predictor, your idea doesn't work; we have to work a bit more.  The problem is you don't know which bit is predictable; it could be any of the bits.  Here's how we handle that.  We prove that if none of the bits is predictable, then there can be no distinguisher.  You prove this using a hybrid argument (do you know what that is?).  Then, take the contrapositive, and you get the result you wanted.