By definition, $\mathcal{A} \otimes \mathcal{B}$ is the $\sigma$-algebra on $X \times Y$ generated by the set $\mathcal{A} \times \mathcal{B}=\{A \times B: A \in \mathcal{A}, B \in \mathcal{B}\}$, (or equivalently the smallest $\sigma$-algebra that makes the projections measurable).
As $\mathcal{E} \subseteq \sigma_X(\mathcal{E}) = \mathcal{A}$ by definition and likewise $\mathcal{F} \subseteq \sigma_Y(\mathcal{F}) = \mathcal{B}$, we have that
$$\mathcal{E} \times \mathcal{F} = \{E \times F: E \in \mathcal{E}, F \in \mathcal{F}\} \subseteq \mathcal{A} \times \mathcal{B}$$
so $$\mathcal{C} = \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) \subseteq \sigma_{X \times Y}(\mathcal{A} \times \mathcal{B}) = \mathcal{A} \otimes \mathcal{B}$$
This uses the obvious fact (by the definitions) that if $\mathcal{G},\mathcal{G}'$ are families of subsets of any set $Z$, then $\mathcal{G} \subseteq \mathcal{G}'$ implies $\sigma_Z(\mathcal{G}) \subseteq \sigma_Z(\mathcal{G}')$ as well.
The simple example in this post shows that we indeed need some condition like $X \in \mathcal{E}$ and $Y \in \mathcal{F}$ to show the reverse inclusion $\mathcal{A} \otimes \mathcal{B} \subseteq \mathcal{C}$ as well. For this inclusion it suffices to show that $$\mathcal{A} \times \mathcal{B} \subseteq \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) = \mathcal{C}\tag{1}$$ and this is more subtle:
Define $$\mathcal{A}' = \{A \subseteq X: (\pi_X)^{-1}[A] \in \mathcal{C}\}$$
where $\pi_X: X \times Y \to X$ is the projection.
It is easy to check that this defines a $\sigma$-algebra on $X$, by the properties of inverse images and the fact that $\mathcal{C}$ is a $\sigma$-algebra. Also, for $E \in \mathcal{E}$ (and because $Y \in \mathcal{F}$), we have that
$$(\pi_X)^{-1}[E] = E \times Y \in \mathcal{E} \times \mathcal{F} \subseteq \mathcal{C}$$
so that $\mathcal{E} \subseteq \mathcal{A}'$ which means that
$\sigma_X(\mathcal{E}) = \mathcal{A} \subseteq \mathcal{A}'$ as well, or equivalently:
$$\forall A \in \mathcal{A}: A \times Y \in \mathcal{C}\tag{2}$$
Using the analogous argument for $\mathcal{B}$ and $\pi_Y$ and the assumption that $X \in \mathcal{E}$ we also get:
$$\forall B \in \mathcal{B}: X \times B \in \mathcal{C}\tag{3}$$
And then note that $(2)$ together with $(3)$ imply $(1)$ by the simple fact that
$$A \times B = (X \times B) \cap (A \times Y)$$
using that $\mathcal{C}$ is closed under intersections.
This concludes the proof of the reverse inclusion.
Best Answer
Your argument for the one inclusion is correct. Let me propose an argument proving both inclusions at once, which might seem a bit abstract, but highlights what's going on. Some details are left for you to check. The point is that there is an obvious bijection $f\colon X\rightarrow\{c\}\times X,\,x\mapsto(c,x)$. This induces a bijection $f_{\ast}\colon\mathcal{P}(X)\rightarrow\mathcal{P}(\{c\}\times X),\,A\mapsto\{c\}\times A$. Note that this bijection preserves the empty set, complementation and unions (and hence also intersections), and containment. From this, it follows that a subset $\mathcal{F}\subseteq\mathcal{P}(X)$ is a $\sigma$-algebra on $X$ if and only if $f_{\ast}(\mathcal{F})\subseteq\mathcal{P}(\{c\}\times X)$ (note that $f_{\ast}(\mathcal{F})$ as I write it is the same thing as what you denoted as $\{c\}\times\mathcal{F}$) is a $\sigma$-algebra on $\{c\}\times X$. Now let $\mathcal{G}\subseteq\mathcal{P(X)}$ be arbitrary. Putting the previous observations together, we obtain \begin{align*} f_{\ast}(\sigma(\mathcal{G}))&=f_{\ast}\left(\bigcap_{\mathcal{F}\subseteq\mathcal{P}(X)\colon\mathcal{G}\subseteq\mathcal{F},\,\mathcal{F}\text{ is a $\sigma$-algebra}}\mathcal{F}\right)\\ &=\bigcap_{\mathcal{F}\subseteq\mathcal{P}(X)\colon\mathcal{G}\subseteq\mathcal{F},\,\mathcal{F}\text{ is a $\sigma$-algebra}}f_{\ast}(\mathcal{F})\\ &=\bigcap_{\mathcal{F}^{\prime}\subseteq\mathcal{P}(\{c\}\times X)\colon f_{\ast}(\mathcal{G})\subseteq\mathcal{F}^{\prime},\,\mathcal{F}^{\prime}\text{ is a $\sigma$-algebra}}\mathcal{F}^{\prime}\\ &=\sigma(f_{\ast}(\mathcal{G})). \end{align*} In fact, the same argument shows that a bijection between two sets allows us to identify their power sets in a nice enough way such that $\sigma$-algebras correspond to $\sigma$-algebras and that the way in which $\sigma$-algebras are generated from subsets correspond to one another as well.
Slightly differently put, we see that your argument is actually all that we need. Indeed, your argument is this setting says that $f_{\ast}(\sigma(\mathcal{G}))$ is a $\sigma$-algebra containing $f_{\ast}(\mathcal{G})$, whence $\sigma(f_{\ast}(\mathcal{G}))\subseteq f_{\ast}(\sigma(\mathcal{G}))$. However, the exact same reasoning also applies for $f^{-1}$ in place of $f$ and $f_{\ast}(\mathcal{G})$ in place of $\mathcal{G}$, which yields the inclusion $\sigma(\mathcal{G})=\sigma((f^{-1})_{\ast}(f_{\ast}(\mathcal{G})))\subseteq(f^{-1})_{\ast}(\sigma(f_{\ast}(\mathcal{G})))$. Here, we use that $f_{\ast}$ and $(f^{-1})_{\ast}$ are inverses. Using this again and applying $f_{\ast}$ to both sides, we obtain $f_{\ast}(\sigma(\mathcal{G}))\subseteq f_{\ast}((f^{-1})_{\ast}(\sigma(f_{\ast}(\mathcal{G}))))=\sigma(f_{\ast}(\mathcal{G}))$, which is the other inclusion. So, really, all that you were missing is that the roles of $X$ and $\{c\}\times X$ in your argument were interchangeable!