Homomorphism between $Q_8$ and $S_4$

Your calculations are good and your proof is right, you could perhaps word it the following way to make the logic crystal clear:

• Let $\phi$ be some arbitrary function from $Q_8$ to $S_4$ that happens to take $i$ to $(1234)$ and $j$ to $(1243)$.

• Let's show that $\phi$ is not a homomorphism.

• If $\phi$ were a homomorphism, then $\phi(i) = \phi(jij) = \phi(j)\phi(i)\phi(j)$. But this implies $$ (1234) = (1243)(1234)(1243) $$ which is false: the right hand side fixes $1$ and the left hand side doesn't.

Your argument is correct. There is another, perhaps simpler argument: $$ (1234)^2=(13)(24),\qquad (1243)^2=(14)(23) $$ so $\phi(i^2)=(\phi(i))^2=(13)(24)$. On the other hand $\phi(j^2)=(14)(23)$. A contradiction, because $i^2=j^2=-1$.