Given the stochastic differential equation:
$dx=-xdt+dW$
and the initial condition $\left( t_{0},x_{0}\right) $, the solution trajectory $x\left( t;t_{0},x_{0}\right) $ can be derived by variation of constants as
$ x\left( t;t_{0},x_{0}\right) =x_{0}e^{-\left( t-t_{0}\right) }+\int _{t_{0}}^{t}e^{-\left( t-s\right) }dW\left( s\right) $
The expectation is shown to be $ E\left[ x\left( t;t_{0},x_{0}\right) \right] =x_{0}e^{-\left( t-t_{0}\right) } $. Given $a>0$, the probability of $x\left( t\right) $ to be in the open set $\left( -a,a\right) $ is
$ P\left[ x\left( t\right) \in\left( -a,a\right) \right] =1-P\left[ \left\vert x\left( t\right) \right\vert \geq a\right] $
By virtue of the Markov inequality $ P\left[ \left\vert x\left( t\right) \right\vert \geq a\right] \leq \frac{\left\vert x_{0}\right\vert e^{-\left( t-t_{0}\right) }}{a} $ and taking the limit as $t\rightarrow\infty$, $P\left[ \left\vert x\left( t\right) \right\vert \geq a\right] $ converges to zero.
My main concern has to do with $x\left( t\right) $. The solution is an Ornstein-Uhlenbeck process and as a consequence it becomes a Gaussian distribution. This indicates that $x\left( t\right) $ exists in any large region on $\mathbb{R}$ with non-zero probability for any $t$. This indicates that the solution cannot converge to a compact set. However $ \lim_{t\rightarrow\infty}P\left[ x\left( t\right) \in\left( -a,a\right) \right] =1 $ that is the trajectory converges to the compact set $\left[ -a,a\right] $. What is wrong in this reasoning?
You have $\mathbb{E}[X(t)] = x_0 e^{-(t-t_0)}$, but you applied Markov's inequality to $|X(t)|$, effectively turning $\mathbb{E}[|X(t)|]$ into $\mathbb{E}[X(t)]$. To exaggerate the mistake, you could have said that if $Z \sim N(0,1)$, then $$P(|Z| \geq a) \leq \frac{E[Z]}{a} = 0.$$