Let $A[T]$ be the polynomial ring over a ring $A$, and $B$ any ring.
Suppose that $\phi:A\rightarrow B$ is a given ring homomorphism; show that ring homomorphisms $\psi:A[T]\rightarrow B$ extending $\phi$ are in one-to-one correspondence with elements in $B$.
I think it is easy to show that for every $b\in B$ there is a homomorphism: Let $f=\sum_{i=0}^na_iT^i$, then define $\phi(T)=b$, so that $$\phi(f)=\sum_{i=0}^n\phi(a_i)b^i$$ Define this extended homomorphism $\psi$.
I'm having trouble showing that any homomorphism $\psi$ can be associated with a $b\in B$, but I have some idea: Again, since $\psi$ is a homomorphism, we have $$\psi(f)=\sum_{i=0}^n\psi(a_i)\psi(T)^i=\sum_{i=0}^n\psi(a_i)b^i$$ It is not clear to me what to do with the $\psi(a_i)$-term here. I guess I don't know a priori that $\phi=\psi\mid_A$.
Any help is much appreciated!
Your proof is correct. You take $b=\psi(T)$.
It is supposed that $\psi$ extends $\phi$, that is $\psi|_A=\phi$, so you're done.