Proving that a trivial product sigma algebra is the product of sigma algebras

Suppose I consider the space {$c$}$\times X$, for some singleton set {$c$} and measurable space $X$, whose sigma-algebra is $\mathcal{F}$ is generated by $\mathcal{G}$, so $\mathcal{F}=\sigma(\mathcal{G})$. I simply want to show that

$\sigma$({{$c$}}$\times\mathcal{G}$)= {{$c$}}$\times \mathcal{F}$,

which doesn't seem like it should be too hard. The inclusion $\subseteq$ holds immediately, since {{$c$}}$\times \mathcal{F}$ is itself a sigma-algebra containing {{$c$}}$\times\mathcal{G}$, and hence $\sigma$({{$c$}}$\times\mathcal{G}$) being the smallest of such sigma-algebras makes $\subseteq$ immediate. But I can't seem to make any headway proving the reverse inclusion. Does anybody have any insights here?

(The notation {{$c$}}$\times\mathcal{G}$ means sets of the form {$c$}$\times E$ for $E \in \mathcal{G}$, and likewise for {{$c$}}$\times \mathcal{F}$.)

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!