Assume you have an 1-dim real-valued SDE like $dX_t=b(t,X_t)dt+\sigma(t,X_t)dB_t$ on a prob. space $(\Omega, \mathcal{F}, \mathcal{F}_t, \mathbb{P})$ with an Brownian motion $B_t$, where path-wise uniqueness holds and there exist for every start condition $X_0$ (independent from $B_t$) a strong solution to the SDE.
Further assume that for $K_1=X_0$ and $K_2=X_0$ you have two constant solutions $X^1_t=K_1$, $X^2_t=K_2$ with $K_1<K_2$ and $K_1,K_2\in \mathbb{R}$. Does it immediately hold that:
if a solution starts in $[K_1,K_2]$; $X_0\in [K_1,K_2]$ $\mathbb{P}$-a.s.
then $\mathbb{P}(X_t\in [K_1,K_2]: \forall t\geq 0)=1$
I guess that is true, if we have $b(t,X_t)=b(X_t)$ and $\sigma(t,X_t)=\sigma(X_t)$ (time-independency of the coefficients), because if $X_t$ would reach the level $K_1$ or $K_2$, we could "start" the solutions at this time point and by path-wise uniqueness and the assumptions of the constant solutions, it must hold that these levels are absorbing.
If a solution is constant, that means $$\int_0^t b(t, X_t) dt = - \int_0^t \sigma(t, X_t) dW_t$$ The left process has quadratic variation 0, and the right one $\int_0^t \sigma^2(s,X_s) ds $, so in particular $\sigma(t, X_t) = 0$ for almost all t. Consequently the same holds for b, and the solution is a ODE which is just $dX_t = 0$, which is only solved by constants. If solution starts in $X_0 \in [K_1, K_2]$, then it remains in that interval for all time as it is constant