*Intuition*

The result seems plausible for the following reason: when $Y_0$ is close to zero, $Y_t$ behaves like the process $\tilde Y_t$ which satisfies the linear SDE: $$
d \tilde Y_t = X_t \tilde Y_t dt + X_t \tilde Y_t d V_t \;.
$$   This linear SDE has the pathwise solution: $$
\tilde Y_t = Y_0 e^{ \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds } \;.
$$ Since almost surely $$
 \frac{1}{t} \left( \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds\right) \to - \frac{1}{4} \quad \text{as $t \to \infty$ } \tag{$\star$}
$$ it follows that almost surely $\tilde Y_t \to 0$ as $t \to \infty$.  A similar argument holds if $Y_0$ is close to one.  

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Recall the SDE $$
d Y_t = Y_t (1- Y_t) X_t (dt + d V_t) \tag{1}
$$ where $V$ is a standard Brownian motion and $X$ is an Ornstein-Uhlenbeck process with stationary distribution $\mathcal{N}(0,1/2)$.  

> Thm. Almost surely $\lim_{t \to \infty} Y_t \in \{0, 1\}$.


*Proof*


The main tool in this proof is the function $f(y) = \log(y) -  \log( 1-y)$ which bijectively maps the unit interval $(0,1)$ to $\mathbb{R}$ by mapping $0$ to $-\infty$ and $1$ to $+\infty$.  

By the change of variables $Z_t = f(Y_t)$, the SDE (1) can be reduced to the following SDE $$
d Z_t = X_t dt - \frac{1}{2} X_t^2 dt + X_t dV_t  + \frac{e^{Z_t}}{e^{Z_t}+1} X_t^2 dt \;. \tag{2}
$$ Set $\tilde Z_t = Z_t -\int_0^t (X_s ds - (1/2) X_s^2 ds + X_s d V_s)$ to get $$
d \tilde Z_t =   \frac{e^{\tilde Z_t +\int_0^t (X_s ds - \frac{1}{2} X_s^2 ds + X_s d V_s)}}{e^{\tilde Z_t +\int_0^t (X_s ds - \frac{1}{2} X_s^2 ds + X_s d V_s)}+1}  X_t^2 dt \;. \tag{3}
$$ Note from (3) that $\tilde Z_t$ is non-decreasing.  Hence, there are two possibilities: either $\lim_{t \to \infty} \tilde Z_t$ is finite or $+\infty$.  If it is finite, then $\lim_{t \to \infty} Z_t = -\infty$ by ($\star$), and otherwise, $\lim_{t \to \infty} Z_t =  +\infty$.  In the original variables, this means that $\lim_{t \to \infty} Y_t = 0$ or $1$ almost surely -- as required.