Given a realization of the Ornstein-Uhlenbeck process $X_t$, the SDE $$ d Y_t = Y_t (1- Y_t) X_t (dt + d V_t) \tag{1} $$ is scalar, nonautonomous, and nonlinear. Note that (1) has two fixed points at $0$ and $1$, which are asymptotically stable in the following sense.
Theorem. For almost all $Y_0 \in (0,1)$, we have $\lim_{t \to \infty} Y_t \in \{0, 1\}$ almost surely.
Intuitive Explanation
This theorem is 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: $$ \tilde Y_t = Y_0 e^{ \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds } \;. $$ By the strong law of large numbers, $$ \lim_{t \to \infty} \frac{1}{t} \left( \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds\right) = - \frac{1}{4} \quad \text{a.s.} \tag{$\star$} $$ It follows that $\tilde Y_t \to 0$ as $t \to \infty$ almost surely. A similar argument holds if $Y_0$ is close to one.
Proof
A key 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 $-\infty$, $1$ to $+\infty$, and $1/2$ to $0$.
By Itô's Lemma, a.s., for all $t \ge 0$, $$ d f(Y_t) = f'(Y_t) Y_t (1-Y_t) X_t (dt + d V_t) + \frac{1}{2} f''(Y_t) Y_t^2 (1-Y_t)^2 X_t^2 dt \;. $$ Since $$ f'(y) y (1-y) = 1 \quad \text{and} \quad \frac{1}{2} f''(y) y^2 (1-y)^2 = -\frac{1}{2} + y $$ we obtain the following SDE for $Z_t = f(Y_t)$ $$ d Z_t = \frac{1}{2} \tanh( \frac{Z_t}{2} ) X_t^2 dt + X_t ( dt + dV_t ) \;. \tag{2} $$$$ d Z_t = \frac{1}{2} \tanh\left( \frac{Z_t}{2} \right) X_t^2 dt + X_t ( dt + dV_t ) \;. \tag{2} $$
Note from (2), and the fact that $-1 < \tanh(x) < 1$ for all $x \in \mathbb{R}$, $$ \int_0^T ( - \frac{1}{2} X_t^2 dt X_t dt + X_t d V_t) < Z_T - Z_0 < \int_0^T ( \frac{1}{2} X_t^2 dt + X_t dt + X_t d V_t) $$$$ \int_0^T \left( - \frac{1}{2} X_t^2 dt + X_t dt + X_t d V_t\right) < Z_T - Z_0 < \int_0^T \left( \frac{1}{2} X_t^2 dt + X_t dt + X_t d V_t\right) $$ and hence, a.s., $$ -\frac{1}{4} < \lim_{T \to \infty} \frac{Z_T}{T} < \frac{1}{4} . \tag{3} $$ Note that (3) limits how fast $Z_T$ can diverge.
Let $z^{(0)} < z^{(1)}$ be two different initial conditions for (2), and let $Z^{(0)}_t$ and $Z^{(1)}_t$ be the corresponding paths emanating from these initial conditions which satisfy: \begin{align*} Z^{(0)}_T &= z_0^{(0)} + \int_0^T \left( \frac{1}{2} \tanh(\frac{Z_t^{(0)}}{2}) X_t^2 dt + X_t ( dt + dV_t ) \right) \;, \\ Z^{(1)}_T &= z_0^{(1)} + \int_0^T \left( \frac{1}{2} \tanh(\frac{Z_t^{(1)}}{2}) X_t^2 dt + X_t ( dt + dV_t ) \right) \;. \\ \end{align*}\begin{align*} Z^{(0)}_T &= z^{(0)} + \int_0^T \left\{ \frac{1}{2} \tanh\left(\frac{Z_t^{(0)}}{2}\right) X_t^2 dt + X_t ( dt + dV_t ) \right\} \;, \\ Z^{(1)}_T &= z^{(1)} + \int_0^T \left\{ \frac{1}{2} \tanh\left(\frac{Z_t^{(1)}}{2}\right) X_t^2 dt + X_t ( dt + dV_t ) \right\} \;. \\ \end{align*} We stress that these paths are driven by the same realization of Brownian motion $V_t$ and Ornstein-Uhlenbeck process $X_t$. Hence, a.s., $$ Z^{(1)}_T - Z^{(0)}_T = z_0^{(1)} - z_0^{(0)} + \frac{1}{2} \int_0^T \left( \tanh(\frac{Z_t^{(1)}}{2}) - \tanh(\frac{Z_t^{(0)}}{2}) \right) X_t^2 dt \;. $$$$ Z^{(1)}_T - Z^{(0)}_T = z^{(1)} - z^{(0)} + \frac{1}{2} \int_0^T \left\{\tanh\left(\frac{Z_t^{(1)}}{2}\right) - \tanh\left(\frac{Z_t^{(0)}}{2}\right) \right\} X_t^2 dt \;. $$ Since $\tanh$ is increasing, the difference $Z^{(1)}_T - Z^{(0)}_T$ itself is increasing and \begin{align*} \lim_{T \to \infty} \frac{Z^{(1)}_T - Z^{(0)}_T}{T} &= \lim_{T \to \infty} \frac{1}{T} \int_0^T \frac{1}{2} \left( \tanh(\frac{Z_t^{(1)}}{2}) - \tanh(\frac{Z_t^{(0)}}{2}) \right) X_t^2 dt \\ &\ge \frac{1}{2} \left( \tanh(\frac{z_0^{(1)}}{2}) - \tanh(\frac{z_0^{(0)}}{2}) \right) \lim_{T \to \infty} \frac{1}{T} \int_0^T X_t^2 dt \\ &\ge \frac{1}{4} \left( \tanh(\frac{z_0^{(1)}}{2}) - \tanh(\frac{z_0^{(0)}}{2}) \right) >0 \tag{4} \end{align*}\begin{align*} \lim_{T \to \infty} \frac{Z^{(1)}_T - Z^{(0)}_T}{T} &= \lim_{T \to \infty} \frac{1}{T} \int_0^T \frac{1}{2} \left\{ \tanh\left(\frac{Z_t^{(1)}}{2}\right) - \tanh\left(\frac{Z_t^{(0)}}{2}\right) \right\} X_t^2 dt \\ &\ge \frac{1}{2} \left\{ \tanh\left(\frac{z^{(1)}}{2}\right) - \tanh\left(\frac{z^{(0)}}{2}\right) \right\} \lim_{T \to \infty} \frac{1}{T} \int_0^T X_t^2 dt \\ &\ge \frac{1}{4} \left\{ \tanh\left(\frac{z^{(1)}}{2}\right) - \tanh\left(\frac{z^{(0)}}{2}\right) \right\} >0 \tag{4} \end{align*} In other words, the difference $Z^{(1)}_T - Z^{(0)}_T$ a.s. diverges as $T \to \infty$.
Now suppose that $$ \lim_{T \to \infty} \frac{Z^{(0)}_T}{T} = 0 \;. $$ This can happen if, e.g., the realization $Z^{(0)}_T$ asymptotes to a finite value or diverges at a sublinear rate. However, this can happen at most once, since for any $z^{(\star)} \ne z^{(0)}$ the previous result in (4) implies that $$ \begin{cases} \lim_{T \to \infty} \frac{Z^{(\star)}_T}{T} < 0 ~~\text{if $z^{(\star)} < z^{(0)}$ }\\ \lim_{T \to \infty} \frac{Z^{(\star)}_T}{T} > 0 ~~\text{if $z^{(\star)} > z^{(0)}$ } \end{cases} $$ where $Z^{(\star)}_T$ is the realization with initial condition $z^{(\star)}$. In other words, realizations corresponding to initial conditions: (i) to the left of $z^{(0)}$ diverge to $-\infty$; and (ii) to the right of $z^{(0)}$ diverge to $+\infty$. Hence, there can be at most one initial condition $z^{(0)}$ such that $\lim_{T \to \infty} \frac{Z^{(0)}_T}{T} = 0$ -- otherwise one gets the contradiction that some realizations diverge to $\pm \infty$ simultaneously.
In the original variables, this implies that: for all, but at most one initial condition, we have $\lim_{t \to \infty} Y_t \in \{0,1 \}$ almost surely -- as required.
