Here's a proof sketch from first principles. This should work even if your professor "skips around".
THEOREM $\ $ TFAE for a field $\rm\:R\:$ and a ring hom \rm\ f\:: R\to R'
$\rm (1)\ \ \ f\:$ is not one-one
$\rm (2) \ \ \ f(r) = 0\ $ for some $\rm\ r\ne 0,\ \ r\in R$
$\rm (3) \ \ \ f(1) = 0$
$\rm (4) \ \ \ f(R) = 0$
Proof $\rm\ (1\Rightarrow 2)\ \ \ a\ne b,\ f(a) = f(b)\ \Rightarrow\ f(a-b) = f(a)- f(b) = 0$
$\rm\ (2\Rightarrow 3)\ \ \ r\ne 0\ \Rightarrow 1/r\in R\ \Rightarrow\ f(1) = f(r\cdot 1/r) = f(r)\ f(1/r) = 0$
$\rm\ (3\Rightarrow 4)\ \ \ f(r) = f(1\cdot r) = f(1)\ f(r) = 0$
$\rm\ (4\Rightarrow 1)\ \ \ R$ a field $\rm\Rightarrow 1\ne 0\:,\:$ so $\rm\ f(1) = f(0) = 0\ \Rightarrow\ f\:$ is not one-one.