Proof check: Filtration $\mathbb F$ is not right-continuous

Question

Question

Let $\Omega=C([0,2])$ (set of continious funtions) and $X$ a stochastic process on $\mathbb R$ such that $$X_t(\omega):=\omega(t)$$ with the natural filtration $\mathbb F:=(\mathcal F_t)_{t\in [0,T]}$ given by $\mathcal F_t:=\sigma(X_s:s\le t)$.

I would like to prove that $\mathbb F$ is not right-continuous.

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Proof Attempt

Let $t>0$ and $S_+^t:=\{\omega \in \Omega:X_t(\omega)>0\}=X^{-1}_t((0,\infty))$, then $S_+^t \in \mathcal F_{0+}$, where $\mathcal F_{0+}:=\bigcap_{\varepsilon>0}\mathcal F_{0+\varepsilon}$, but $S_+^t$ is not $\mathcal F_0$-measurable since $t>0$.

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Is this proof correct?

Answer 1

As discussed in the comments, if we define $\bar S := \bigcap_{t > 0} S^t_+$, then $\bar S \in \mathcal F_{0+}$. The last step is to verify that $\bar S \not \in \mathcal F_0$. Since $\mathcal F_0 = \sigma(X_0)$, it is enough to show that there exist $\omega_1, \omega_2 \in \Omega = C([0,2])$ such that $\omega_1(0) = \omega_2(0)$ and $\omega_1 \in \bar S$ but $\omega_2 \not \in \bar S$. So we can just let $\omega_1(t) := t$ and $\omega_2(t) := -t$.