Approximating indicator function on open set continuous functions

Given some polish space $X$ and a probability measure $P$ on $X$ is it true that for any open set $A$ in $X$ one can construct a sequence of bounded continuous functions $\{f_{n}(x)\}_{n}$ such that $f_{n}$ converges in probability to the indicator function on $A$, $I_{A}$?

Yes. We can choose a compatible metric $$d$$ on $$X$$ such that $$0 \leq d \leq 1$$.
For a non-empty set $$C \subset X$$ let $$d(x,C) = \inf_{c \in C} d(x,c).$$ If $$C$$ is empty, set $$d(x,C) = 1$$ for all $$x$$. Then $$x \mapsto d(x,C)$$ is continuous since $$\lvert d(x,C) - d(y,C)\rvert \leq d(x,y)$$.
Assume that $$A$$ is a proper open subset. Let $$F_n = \{x \in X \mid d(x, X \setminus A) \geq \frac1n\}$$. Then $$F_n \subseteq F_{n+1} \subseteq A$$ and $$A = \bigcup_{n=1}^{\infty} F_n$$. The continuous functions $$f_n(x) = \frac{d(x,X\setminus A)}{d(x,X \setminus A)+d(x,F_n)}$$ satisfy $$0 \leq f_n \leq 1$$ and converge point-wise everywhere and monotonically to $$I_A$$.
In particular, $$f_n \to I_A$$ in probability for every probability measure $$P$$, thus answering the OP's question.
Notice that $$f_n(x) = 0$$ for $$x \in X \setminus A$$ and that $$f_n(x) = 1$$ if and only if $$x \in F_n$$.