First let $t$ be fixed and look at the set $\{ w \mid w(t) \in U\}$. Take a sequence of continuous functions $f_k$ that are uniformly bounded and converge to the indicator function of a set $U$ in the Borel-Algebra of $\mathbb{R}^n$. Let $\pi_t$ be the projection on the $t$-th component of Omega. Then $$\int_\Omega (\pi_t)^*f_k d \mu = \int_{\mathbb{R}^n} p(t, x, y) f_k(y) d y$$ by construction. Because the $f_k$ are bounded and the measure is finite, they are dominated by an integrable function and by Lebesgue's theorem on bounded convergence, $$\mu(\{w(t) \in U\}) = \lim_{k \rightarrow \infty}\int_\Omega (\pi_t)^*f_k d \mu = \int_{\mathbb{R}^n} p(t, x, y) \lim_{k \rightarrow \infty} f_k(y) d y = \int_{U} p(t, x, y) d y $$ Hence non-continuity is no real problem.
To now tackle your problem, notice that the indicator functions of the cylinder sets $$ G_{N} := \{ w \mid w(k/N) \in U \forall k = 1, \dots, N^2 \}$$ converges a.s. to the indicator function of $G := \{ w \mid w(t) \in U \forall t \geq 0 \}$ as $N \longrightarrow \infty$: Clearly $G \subset G_N$. On the other hand, if $w \notin A$, then $w(q) \notin U$ for some rational number $q$ and therefore $w \notin G_N$ for $N$ big enough.
The dominated convergence theorem gives then $$ \mu(G) = \lim_{N \rightarrow \infty} \int_U \dots \int_U \left(\prod_{k=1}^{N^2} p(1/N, x_{k-1}, x_k) \right) d x_1 \dots d x_{N^2}$$ If this is helpful at all, I cannot say.