Non-Borel set in arbitrary metric space Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum. 
 A: Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $\mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(\{x,y\},d)$ equipped with the discrete metric $d:\{x,y\}\times \{x,y\} \to \{0,1\}$ given by
$$
d(x,y)=1, \quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
\{\{x\},\{y\},\{x,y\},\emptyset\} = \mathcal{P}(\{x,y\})
$$
where $\mathcal{P}(\{x,y\})$ is the powerset of $\{x,y\}$, so all subsets are Borel measurable sets.
A: Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
