This is not true in general. The following is a counterexample:
For each $n > 0$, let $[n] = \{1, \cdots, n\}$. Fix a sequence $(n_k)_{k \in \mathbb{N}}$ s.t. $\prod_{k \in \mathbb{N}} \frac{n_k - 1}{n_k} = \varepsilon > 0$. Let $X = [2] \times \prod_{k \in \mathbb{N}} [n_k]$ be equipped with the product topology and the product $\mu$ of normalized counting measures. We observe that $X$ is compact and metrizable, so we may regard it as a compact metric space. $\mu$ is easily seen to be a Borel probability measure. For notational purposes, we shall call the $[2]$ coordinate the zeroth coordinate and the $[n_k]$ coordinate the $k$-th coordinate. Let $f: X \rightarrow [2] = \{1, 2\}$ be the projection onto the zeroth coordinate. $f$ is obviously continuous.
Now, we construct $A_n$ as follows: It is generated by the following partition of $X$ into clopen subsets,
$$\left\{[2] \times \{p\} \times \prod_{k > n} [n_k]: p \in \prod_{k = 1}^n [n_k], \textrm{no coordinate of }p\textrm{ is }1\right\} \cup \left\{\{1\} \times \{p\} \times \prod_{k > n} [n_k], \{2\} \times \{p\} \times \prod_{k > n} [n_k]: p \in \prod_{k = 1}^n [n_k], \textrm{ some coordinate of }p\textrm{ is }1\right\}$$
We easily observe that these partitions grow finer as $n$ increases, whence $A_n$ is an increasing sequence of $\sigma$-algebras. One may calculate that,
$$\mathbb{E}(f\mid A_n)(p) = \begin{cases} f(p), & \textrm{if some coordinate between the first and the }n\textrm{-th place is }1\\
\frac{3}{2}, & \textrm{otherwise}
\end{cases}$$
Since $\mathbb{E}(f\mid A_n) \rightarrow \mathbb{E}(f\mid A)$, we see that,
$$\mathbb{E}(f\mid A)(p) = \begin{cases} f(p), & \textrm{if some non-zeroth coordinate of } p \textrm{ is }1\\
\frac{3}{2}, & \textrm{otherwise}
\end{cases}$$
Let $F = \mathbb{E}(f\mid A)$. Since $\textrm{range}(f) = \{1, 2\}$, if $F$ is a.e. equal to a continuous function, we must in particular have $F^{-1}(\frac{3}{2})$ is up to a null set clopen. However, $F^{-1}(\frac{3}{2})$ is given by
$$F^{-1}\left(\frac{3}{2}\right) = [2] \times \prod_{k \in \mathbb{N}} ([n_k] \setminus \{1\})$$
This set is closed, with empty interior (since any open set must contain a basic open set, and by definition of the product topology any basic open set can only place constraints on finitely many coordinates), and has measure $\prod_{k \in \mathbb{N}} \frac{n_k - 1}{n_k} = \varepsilon > 0$. If $O$ is an open set that coincides with $F^{-1}(\frac{3}{2})$ up to a null set, then $O \setminus F^{-1}(\frac{3}{2})$ is an open set with zero measure. Applying the same arguments showing $F^{-1}(\frac{3}{2})$ has empty interior, we see that the only open set with zero measure is the empty set. Hence, $O \setminus F^{-1}(\frac{3}{2}) = \varnothing$, i.e., $O \subseteq F^{-1}(\frac{3}{2})$. But the latter set has no interior, so $O = \varnothing$. But $O$ coincides with $F^{-1}(\frac{3}{2})$ up to a null set, so in particular $\mu(O) = \mu(F^{-1}(\frac{3}{2})) = \varepsilon > 0$, a contradiction.
