Let $K$ be a field, $L/K$ ad $F/K$ field (algebraic) extensions, and let $\sigma:FL \longrightarrow \overline{K}$ be a field homomorphism (so is injective), where $\overline{K}$ is the algebraic closure of $K$ and $FL$ is the composite of $F$ and $L$ (the smaller subfield of $\overline{K}$ that contains both $F$ and $L$). How can I prove that $\sigma(FL)=\sigma(F)\sigma(L)$?
This is what I did: since $F,L \subseteq FL$ we have that $\sigma(F),\sigma(L) \subseteq \sigma (FL)$, so this means that, for the minimality of $\sigma(F)\sigma(L)$, we have that $\sigma(F)\sigma(L) \subseteq\sigma(FL)$.
Is this correct? How can I do the other inclusion?
This follows easily from the description via elements: $$\textstyle F \cdot L = \{\sum_i a_i b_i : a_i \in F, \, b_i \in L\}$$ But you can also prove it with the definition as the smallest subfield that contains both fields:
You have already proved $\sigma(F) \sigma(L) \subseteq \sigma(FL)$. Conversely, observe that $E := \sigma^{-1}(\sigma(F)\sigma(L))$ is a subfield of $FL$, and that $F \subseteq \sigma^{-1}(\sigma(F)) \subseteq E$ and similarly $L \subseteq E$. Hence, $FL \subseteq E$. This means $\sigma(FL) \subseteq \sigma(F) \sigma(L)$.