Here is a theorem from Judson and part of its proof:
I am concerned about the part that is squared in red. Why do we have to show that the function is injective? Is this a common tactic in finding an extension field? Why is the image of $\psi$ important?
So the proof sets up $\psi:F \rightarrow F[x]/\langle p(x)\rangle \equiv E$. We prove then that $\psi$ is structure preserving and injective. This shows that $F$ has an inclusion via $\psi$ into $E$, i.e. we can identify $F$ with a subfield of $E$, which is exactly what it means for $E$ to be a field extension of $F$.
What you are seeing here is exactly analogous to a technique in group theory. To show that a group $G$ can be viewed as a subgroup of some other group $G'$, you simply write down an injective homomorphism $\phi:G\rightarrow G'$.
It is necessary that $\psi$ be injective so that we can be assured nothing is "lost" in the process of trying to include $F$ into $E$. For example, if we dropped the injectivity requirement, every field could then be viewed as a subfield of any other field via the structure preserving map $F \rightarrow 0 \subset E$, for any fields $F$ and $E$.
Another way to see it is that since $\psi$ is injective, it is isomorphic onto its image, and thus $F$ is isomorphic to a subfield of $E$.