I am looking at an SDE of the form $d{X_t} = \left( {{1_A}({X_t}) - {1_{{A^c}}}({X_t})} \right)d{W_t}$ such that ${X_0} = 0$, $A \in \mathcal{B}(\mathbb{R})$ and ${A^c}$ has a lebesgue measure of zero. I was thinking along the following lines: Consider a complete probability space $(\Omega ,\mathcal{F},\mathbb{P})$ that supports some Brownian Motion $X_t$. Then:
${W_t} = \int\limits_0^t {\left( {{1_A}({X_u}) - {1_{{A^c}}}({X_u})} \right)d{X_u}} $
is a brownian motion by Levy's Characterisation Theorem. Then:
$\left( {{1_A}({X_t}) - {1_{{A^c}}}({X_t})} \right)d{W_t} = \left( {{1_A}({X_t}) - {1_{{A^c}}}({X_t})} \right)\left( {{1_A}({X_t}) - {1_{{A^c}}}({X_t})} \right)d{X_t} = d{X_t}$
From which we may conclude that $\left( {X,W} \right)$, $(\Omega ,\mathcal{F},\mathbb{P})$ and $\mathcal{F}_t$ is a weak solution to the above SDE.
Also can we conclude $\mathbb{P}({X_t} \in {A^c}) = 0$ for all t?
Since $X$ is a Brownian Motion, $X_t$ is normally distributed for all $t>0$ and so the fact $\mathbb{P}(X_t \in A^c) = 0$ follows from the fact that $A^c$ has Lebesgue measure $0$. For $t=0$, we have $\mathbb{P}(X_t \in A^c) = 0$ if and only if $0 \in A$.
On the other hand, we don't necessarily have $\mathbb{P}(X_t \in A \, \forall \, t>0) = 1$. In fact, as long as $A^c$ is non-empty, $\mathbb{P}(X_t \in A \, \forall \, t>0) = 0$ because Brownian motion almost surely hits all points in $\mathbb{R}$.