Lemma: Let $f:[0,\infty) \to \mathbb{R}$ be left-continuous on $(0,\infty)$ and with right-hand limits on $[0,\infty)$. Then: $f$ is continuous on $[0,t_0)$ if, and only if, $$\forall \epsilon>0 \exists \delta>0: |f(s)-f(r)| \leq \epsilon \quad \text{for all} \, |s-r| \leq \delta, s,r \in [0,t_0) \cap \mathbb{Q}. \tag{1}$$
Proof: "$\Rightarrow$": Obvious. "$\Leftarrow$": Since $f$ is left-continuous and has right-hand limits, $f$ is continuous at $t$ if $$f(t) = f(t+) = \lim_{s \downarrow t} f(s).$$ It follows from $(1)$ and the left-continuity that $$\lim_{s \downarrow t, s\in \mathbb{Q}} f(s)= f(t).$$ Since $$\lim_{s \downarrow t} f(s) = \lim_{s \downarrow t, s \in \mathbb{Q}} f(s)$$ the claim follows.
If we rephrase $(1)$ for the stochastic process $X$, then it becomes
$$A_1 := \{X \, \text{continuous on $[0,t_0)$}\} = \bigcap_{\ell>1} \bigcup_{k \geq 1} \bigcap_{\substack{|s-r| \leq \frac{1}{k} \\ s,r \in \mathbb{Q} \cap [0,t_0)}} \left\{|X_s-X_r| \leq \frac{1}{\ell} \right\}.$$
Since $X$ is adapted, we get $A_1 \in \mathcal{F}_{t_0}$ (as a countable union of measurable sets).
Note however, that $(1)$ does not give continuty at $t=t_0$; that is, we need an extra condition to ensure continuity at $t=t_0$ - and here comes the right-continuity of the filtration into play.
$f$ is continuous in $t=t_0$ if and only if for any $\gamma>0$ and $\epsilon>0$ we can find $\delta>0$ such that $$|f(t_0)-f(s)| \leq \epsilon \quad \text{for all $s \in [t_0-\delta,t_0+(\delta \wedge \gamma)] \cap \mathbb{Q}$.}$$
Roughly speaking: In order to decide whether $f$ is continuous at $t=t_0$ it does not suffices to know $f$ up to time $t_0$, we need some information about the future. Consequently,
$$A_2 := \{X \, \text{continuous at $t=t_0$}\} = \bigcap_{m \geq 1} \underbrace{\bigcap_{\ell \geq 1} \bigcap_{\substack{|s-t_0| \leq \frac{1}{k} \\ s \in \mathbb{Q} \cap [0,t_0+\frac{1}{m})}} \left\{|X_s-X_r| \leq \frac{1}{\ell} \right\}}_{\in \mathcal{F}_{t_0+1/m}} \in \mathcal{F}_{t_0+} = \mathcal{F}_{t_0}.$$
Hence, $A= A_1 \cap A_2 \in \mathcal{F}_{t_0}.$