 Solution

Give the step function one private previous-value cell, freshly allocated for each independent Fibonacci generator:

def fibonacci(): Iter[BigInt] = {
  var previous = BigInt(1)
  val step: BigInt => BigInt = { current =>
    val following = previous + current
    previous = current
    following
  }
  new FunIter[BigInt](BigInt(0), step)
}

The initial artificial previous value is $F_{-1}=1$, making the first update return one. Thereafter, just before advancing from $F_n$, the captured cell holds $F_{n-1}$; the assignment produces $F_{n+1}$ and retains $F_n$. Induction gives $0,1,1,2,3,5,8,\dots$. BigInt avoids the fixed-width overflow that would prevent Int from representing all Fibonacci numbers.

Copies made through part (c) share memoized future cells, so interleaving their traversal does not advance this private state twice for the same sequence position. A separate call to fibonacci() receives a separate closure. Another valid construction uses a pure step on the pair $(F_n,F_{n+1})$ and projects the first coordinate; the stateful construction above follows the suggested single-current-value approach.