RSA Blind Signature Attack
You are correct; whoever put together the above proof typo'ed that point; we have $c \times d \equiv 1 \pmod {\phi(n)}$, or more accurately, $c \times d \equiv 1 \pmod {lcm(p-1, q-1)}$.

On the other hand, the attacker isn't expected to be able to compute step 2 (he can't, he doesn't know the value of $lcm(p-1,q-1)$, and hence cannot compute $d$).  Instead, it is there to establish the relationship between $c$ and $d$ (and why they cancel out in step 3)