We want to prove that if $X_t$ is an $\mathcal{F}_t$ - martingale: $\mathbb{E}[X_t | \mathcal{F}_s] = X_s$ for $s<t$, then it's also a $\sigma(\mathcal{F}_s \cup \mathcal{G}_s)$- martingale. $\mathcal{F}_t \perp \mathcal{G}_t $ are filtrations.
The proof I was given involves choosing $F \in \mathcal{F}_t$ and $G \in \mathcal{G}_t$ and proving that
\begin{equation} \int_{F \cap G} \mathbb{E}[X_t | \mathcal{F}_s] dP = \int_{F \cap G} X_s dP \end{equation} and then claiming that $\sigma( \{F \cap G :F \in \mathcal{F}_t, G \in \mathcal{G}_t \} ) = \sigma(\mathcal{F}_t \cup \mathcal{G}_t)$. $\square$
I thought of a different (and more elegant) way, using the Tower property of expectations. Since we know that $\mathcal{F}_t \subset \sigma(\mathcal{F}_s \cup \mathcal{G}_s)$, then
\begin{align} \mathbb{E}[X_t | \sigma(\mathcal{F}_s \cup \mathcal{G}_s)] &=\mathbb{E}[ \mathbb{E}[X_t | \sigma(\mathcal{F}_s \cup \mathcal{G}_s)] | \mathcal{F}_s] \\ &= \mathbb{E}[X_t | \mathcal{F}_s]\\ &= X_s \space \square \\ \end{align}
is my reasoning correct?
Since $\mathcal{F}_s\subseteq\sigma(\mathcal{F}_s\cup\mathcal{G}_s)$ you can conclude that $$ {\rm E}[{\rm E}[X_t\mid\sigma(\mathcal{F}_s\cup\mathcal{G}_s)]\mid\mathcal{F}_s]={\rm E}[X_t\mid\mathcal{F}_s] $$ but you certainly cannot conclude that $$ {\rm E}[{\rm E}[X_t\mid\sigma(\mathcal{F}_s\cup\mathcal{G}_s)]\mid\mathcal{F}_s]={\rm E}[X_t\mid\sigma(\mathcal{F}_s\cup\mathcal{G}_s)] $$ which is what you did in your first equality. This would require that $\sigma(\mathcal{F}_s\cup\mathcal{G}_s)\subseteq \mathcal{F}_s$ which in general does not hold.