Non-Borel set in arbitrary metric space

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.

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.