For a Borel map $f$ from a Borel space $(\Omega, S)$ into a separable metric space $(X, d)$, we need to construct a sequence of Borel maps $\{f_n\}$ such that for every $n, f_n$ can only take countable many values and $\{f_n\}$ converges to $f$ uniformly.
I think I have a valid strategy to prove this, but I can't figure the last step:
To do this, we start by writing $X$ as a countable union of open sets $X = \cup_iB_i$. This is possible since $(X, d)$ is a separable metric space and hence is second countable. Since the Borel $\sigma-$algebra on $X$ is generated by open sets, we can write $X = \cup_iB_i = \cup_iA_i$, where $A_i$'s are disjoint Borel sets. Then, $\Omega = \cup_if^{-1}(A_i)$, where $f^{-1}(A_i)$'s are disjoint Borel subsets of $\Omega$.
So, we can construct a sequence $\{f_n\}$ such that for every $n$, $f_n(\omega) = x_{i_n}$ is same for every $\omega \in f^{-1}(A_i)$. Thus, each $f_n$ can only take countably many values.
However, I cannot figure out how to define $f_n(\omega)$ so that $f_n$ converges to $f$ uniformly.
Let $(x_i)$ be a counatble dense set. Then $X=\bigcup_n B(x_i,\frac 1 N)$ for each $n$. Considering the sets $ B(x_i,\frac 1 N) \setminus \bigcup_{j \leq i} B(x_j,\frac 1 N)$ we see that $X$ is a countable union of disjoint sets with diameter at most $\frac 2 N$. Now take these set as $A_i$'s in your argument. You see that $|f_n(\omega)-f(\omega)| \leq \frac 2 N$.