Let $W$ be a one-dimensional Brownian motion. Consider the stochastic differential equation (SDE)
$$dX_t = C(t)(1-X_t)dW_t,\quad \forall t\ge 0,$$
where $C$ is a continuous and bounded function. Under which condition (on $C$),
If $X_0$ and $C$ are deterministic (or independent of $W$), then $$X_t = 1-(1-X_0)\exp\left(-\int_0^tC(s)\mathrm dW_s-\frac12\int_0^tC(s)^2\mathrm ds\right).$$ This will go to 1 if the argument of the exponential goes to $-\infty$. I claim that this is equivalent to the condition $I:=\int_0^\infty C(s)^2\mathrm ds=+\infty$.
If $I<\infty$, then the argument of the exponential converges in $\mathrm L^2$ to the same integral for $t=\infty$, which has distribution $\mathcal N(-I/2,I)$; this makes it impossible for it to converge almost surely to $-\infty$.
If $I=\infty$, then $M:t\mapsto\int_0^t C(s)\mathrm dW_s$ is a continuous martingale. By the Dambis-Dubins-Schwarz theorem [RY, Theorem V-(1.6) in my edition], there exists a Brownian motion $B$ on the same probability space such that $M_t=B_{\langle M,M\rangle_t}$ for all $t\geq0$. In these terms, the argument of the exponential is $B_{\langle M,M\rangle_t}-\langle M,M\rangle_t/2$. Since almost surely we have $\lim_{\tau\to\infty}(B_\tau-\tau/2)=-\infty$ and $\lim_{t\to\infty}\langle M,M\rangle_t=I=+\infty$, the exponential goes to zero and $X$ goes to 1 almost surely.
[RY] D. Revuz and M. Yor, Continuous Martingales and Brownian Motion, third edition (1993).
