Your second equality in the definition of $F^X_{t_0}$ is incorrect - $F^X_{t_0}$ is the $\sigma$-field generated by events of the form $\{X_s\in B\}$ for $0\le s\le t_0$ and $B\in\mathcal B(\mathbb R)$, but it contains many more events, too. For example, $\{X_{s_1}\in B_1,X_{s_2}\in B_2,\ldots,X_{s_n}\in B_n\}\in F^X_{t_0}$ for all $s_1,\ldots,s_n\in[0,t_0]$ and $B_1,\ldots,B_n\in\mathcal B(\mathbb R)$. Or we could take unions instead, such as $\bigcup_{i=1}^\infty\{X_{s_i}\in B_i\}$. Or we could take functions of multiple $X_s$'s, i.e. something like $\{X_s+X_t\in B\}$.
Informally, an event is $F^X_{t_0}$-measurable if we know whether or not it occurs after observing the process up to time $t_0$. There is a small caveat: we can only make countable many choices. An event of the form $\{X_s\in B_s\text{ for }s\in[0,t_0]\}$, where $B_s\in\mathcal B(\mathbb R)$ for all $s$, is not measurable. However, as long as we stick to countable combinations, we can do whatever we like.
As for what $A$ represents: we already know that the sample paths of $X$ are RCLL, and $A$ is the event that up to time $t_0$, all left and right limits coincide. Specifically,
\begin{align*}
A &=\{\omega\in\Omega:t\mapsto X_t(\omega)\text{ is continuous on }[0,t_0)\}\\
&=\{\omega\in\Omega:\lim_{s\uparrow t}X_s(\omega)=\lim_{s\downarrow t}X_s(\omega)\text{ for all }t\in(0,t_0)\}.
\end{align*}
Using our heuristic description of $F^X_{t_0}$, it seems clear that $A\in F^X_{t_0}$: if you observe the process up to time $t_0$, you know whether or not it was continuous, right? Well, we have to be careful. If $X$ was a process with arbitrary sample paths rather than RCLL, then in fact we would have a negative answer. The reason is we need to know that $X$ is continuous at every point, and as there are uncountably many points, this is too much information for this $\sigma$-field to be able to track.
However, in the RCLL case, things become a little easier. RCLL functions cannot have removable discontinuities, so if the restriction of $X$ to a dense subset of $[0,t]$ was (uniformly) continuous with $t<t_0$, then the function on all of $[0,t]$ must be as well, and then we can let $t\uparrow t_0$ in an appropriate way. In particular, using the fact that $\mathbb Q$ is a countable dense subset of $\mathbb R$, we find that
\begin{align*}
A&=\bigcap_{t\in(0,t_0)\cap\mathbb Q}\{\omega\in\Omega:\forall \varepsilon\in(0,\infty)\cap\mathbb Q,\,\exists\delta\in(0,\infty)\cap\mathbb Q\text{ such that }\\
&\qquad\qquad\qquad\forall u\in[0,t]\cap\mathbb Q,\forall s\in(u-\delta,u+\delta)\cap\mathbb Q\cap[0,t],\,|X_s(\omega)-X_u(\omega)|<\varepsilon\}\\
&=\bigcap_{t\in(0,t_0)\cap\mathbb Q}\bigcap_{\varepsilon\in(0,\infty)\cap\mathbb Q}\bigcup_{\delta\in(0,\infty)\cap\mathbb Q}\bigcap_{u\in[0,t]\cap\mathbb Q}\bigcap_{s\in(u-\delta,u+\delta)\cap\mathbb Q\cap[0,t]}\{|X_s-X_u|<\varepsilon\}
\end{align*}
which is $F^X_{t_0}$-measurable.
(Note that this has been edited to reflect comments by John Dawkins.)
