Prove that $\inf\left\{t\ge 0:Y_t=y\right\}$ is a stopping time

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
• $(Y_t)_{t\ge 0}$ be a real-valued $\mathcal F$-adapted continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$
• $y\in\mathbb R$

Question:

1. How can we show that $$\tau:=\inf\left\{t\ge 0:Y_t=y\right\}$$ is an $\mathcal F$-stopping time?
2. Do we need the continuity of $Y$? Or is it sufficent, if this property holds only $\operatorname P$-almost surely? What about ($\operatorname P$-almost sure) right-continuity?

I'll try to give an answer to 1. by myself. Step 1:

• Let $T\ge 0$ and $$\omega\in\left\{\sup_{t\in[0\:T]}Y_t\ge y\right\}$$
• $Y$ is continuous $\Rightarrow$ $\exists t_{\text{max}}\in[0,T]$ with $$Y_{t_{\text{max}}}(\omega)=\sup_{t\in[0,\:T]}Y_t(\omega)\in[y,\infty)$$ and $$y\in\left[Y_0(\omega),Y_{t_{\text{max}}(\omega)}\right]$$ and hence $\exists t\in[0,T]$ with $$Y_t(\omega)=y$$ by the intermediate value theorem

Step 2:

• Let $T\ge 0$ and $$\omega\in\left\{\tau_y\le T\right\}=\left\{\exists t\in[0,T]:Y_t=y\right\}$$
• It's obvious that then $$\omega\in\left\{\sup_{t\in[0\:T]}Y_t\ge y\right\}$$

Step 3:

• We have shown that $$\left\{\tau_y\le T\right\}=\left\{\sup_{t\in[0\:T]}Y_t\ge y\right\}$$ for all $T\ge 0$
• Since $Y$ is continuous, $$\sup_{t\in[0\:T]}Y_t=\sup_{t\in[0\:T]\cap\mathbb Q}Y_t$$ and since $Y$ is $\mathcal F$-adapted, $$\left\{\sup_{t\in[0\:T]\cap\mathbb Q}Y_t\ge y\right\}\in\mathcal F_T$$
• Thus, we're done

Is the proof correct? If so, I think that at least this proof won't work unless $Y$ is continuous. However, it should work in the case of $\operatorname P$-almost sure continuity too, if we assume that $\mathcal F$ is $\operatorname P$-complete, right?