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Here's an explicit construction that gives a counterexample. For simplicity let $c=0$ (not important).

First, let $\alpha>0$ and consider the probability that a standard Brownian motion started at 0 hits 0 at some time in the interval $(\alpha t, \alpha t+t)$. Then (1) this probability does not depend on $t$ (by Brownian scaling) -- call it $p_\alpha$; and (2) $p_\alpha\to 1$ as $\alpha\to 0$ (because with probability 1, the standard Brownian motion hits 0 at some time in the interval $(0,t)$).

Now we'll use this and Borel-Cantelli to show that with positive probability, we can construct $\sigma$ and a sequence of times $t_n\uparrow 1$ such that $Y_{t_n}=1$ for all $n$.

Let $\alpha_n$ be some sequence decreasing to 0 quickly enough that $\sum (1-p_{\alpha_n}) <\infty$.

Let $t_0=0$ and recursively define $t_1, t_2, \dots$ as follows.

Given $t_n<1$ and $Y(t_n)=0$, let $\sigma_t=\alpha_n$ \sigma_t=\sqrt{\alpha_n}$for$t\in(t_n, (1+t_n)/2)$, and let$\sigma_t=1$for$t\in((1+t_n)/2, t_{n+1})$, where$t_{n+1}$is defined by$t_{n+1}=\inf \big[ t>(1+t_n)/2: Y_t=0 \big]$. The idea of this definition: given$t_n<1$, we divide the remaining time interval$(t_n, 1)$into two halves, and run BM at speed$\alpha_n$on the first half and at speed 1 on the second half, stopping as soon as we hit 0 during the second half. Since we start at 0, the probability that we DO hit 0 at some point during the second half is easily seen to be$p_{\alpha_n}$as defined above. Hence$P(t_{n+1}<1 | t_n<1)=p_{\alpha_n}$. Now using Borel-Cantelli (and reasoning straightforwardly about independence) we get that there is positive probability that$t_n<1$for all$n$. In that case also$t_n\uparrow 1$(since$1-t_{n+1}<(1-t_{n})/2$). Also$Y_{t_n}=0$for all$n$by construction. But the process$Y_t$is continuous (since$\sigma_t$is bounded). So then also$Y_1=0$as desired. 1 Here's an explicit construction that gives a counterexample. For simplicity let$c=0$(not important). First, let$\alpha>0$and consider the probability that a standard Brownian motion started at 0 hits 0 at some time in the interval$(\alpha t, \alpha t+t)$. Then (1) this probability does not depend on$t$(by Brownian scaling) -- call it$p_\alpha$; and (2)$p_\alpha\to 1$as$\alpha\to 0$(because with probability 1, the standard Brownian motion hits 0 at some time in the interval$(0,t)$). Now we'll use this and Borel-Cantelli to show that with positive probability, we can construct$\sigma$and a sequence of times$t_n\uparrow 1$such that$Y_{t_n}=1$for all$n$. Let$\alpha_n$be some sequence decreasing to 0 quickly enough that$\sum (1-p_{\alpha_n}) <\infty$. Let$t_0=0$and recursively define$t_1, t_2, \dots$as follows. Given$t_n<1$and$Y(t_n)=0$, let$\sigma_t=\alpha_n$for$t\in(t_n, (1+t_n)/2)$, and let$\sigma_t=1$for$t\in((1+t_n)/2, t_{n+1})$, where$t_{n+1}$is defined by$t_{n+1}=\inf \big[ t>(1+t_n)/2: Y_t=0 \big]$. The idea of this definition: given$t_n<1$, we divide the remaining time interval$(t_n, 1)$into two halves, and run BM at speed$\alpha_n$on the first half and at speed 1 on the second half, stopping as soon as we hit 0 during the second half. Since we start at 0, the probability that we DO hit 0 at some point during the second half is easily seen to be$p_{\alpha_n}$as defined above. Hence$P(t_{n+1}<1 | t_n<1)=p_{\alpha_n}$. Now using Borel-Cantelli (and reasoning straightforwardly about independence) we get that there is positive probability that$t_n<1$for all$n$. In that case also$t_n\uparrow 1$(since$1-t_{n+1}<(1-t_{n})/2$). Also$Y_{t_n}=0$for all$n$by construction. But the process$Y_t$is continuous (since$\sigma_t$is bounded). So then also$Y_1=0\$ as desired.