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 $\sigma_t=\sqrt{\alpha_n}$
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.