Separability of Reproducing Kernel Hilbert Space of a positive definite covariance function.

I am reading the book Random Fields and Geometry by R. Adler and J. Taylor. In chapter 3 they introduce the concept of reproducing kernel Hilbert space (RKHS). I will summarize the definition in what follows.

Firstly let us consider a Gaussian process $$f$$ defined on $$T$$ a metric space which we assume compact in the canonical metric $$d$$ generated by the process, where $$d(s, t) = \sqrt{\mathbb{E}[(f(s)-f(t))^2]}$$ for $$s,t \in T$$. Let $$C(s,t)$$ be the covariance function of $$f$$ and let $$C$$ be continuous and positive definite. Let us define $$S = \left\{ u \colon \ T \rightarrow \mathbb{R} \colon u(\cdot) = \sum_{n=1}^n a_i C (s_i, \cdot), a_i \in \mathbb{R}, s_i \in T, n \geq 1 \right \}.$$ Define an inner product on $$S$$ by $$(u,v)_H = \left( \sum_{i=1}^n a_i C(s_i, \cdot), \sum_{j=1}^m b_j C(t_j, \cdot)\right)_H.$$ Note that, since we assumed that $$C$$ is positive definite $$(\cdot, \cdot)_H$$ induces a norm. Moreover the reproducing kernel property holds, namely $$(u, C(t, \cdot)_H = \left( \sum_{i=1}^n a_i C(s_i, \cdot), C(t,\cdot)\right)_H = \sum_{i=1}^n a_i C(s_i, t)=u(t).$$ Let us define $$H(C)$$ the reproducing kernel hilbert space of $$f$$ the closure of $$S$$ under the norm induced by the inner product $$( \cdot, \cdot)_H$$.

The the authors state that since $$T$$ is separable (it is a compact metric space) then $$H(C)$$ is separable due to the continuity of $$C$$, how does if follows? My first try was: let $$\{\phi_n\}_{n \in \mathbb{N}}$$ be the orthonormal sequence on $$T$$, then I would like to say that $$\psi_n(\cdot) = C (\phi_n, \cdot)$$ is an orthonormal sequence on $$H(C)$$, but here I stop. Moreover is the inner product unique?

From the reproducing property of the RKHS it follows that the canonical feature map $$\Phi:T \to H(C):t \mapsto C(\cdot ,t)$$ is continuous, since if $$t_n \to t$$ in $$T$$, then $$||C(\cdot,t_n)-C(\cdot,t)||_H^2 = \underbrace{(C(\cdot,t_n),C(\cdot,t_n))_H}_{C(t_n,t_n)}- 2\underbrace{(C(\cdot,t_n),C(\cdot,t))_H}_{C(t_n,t)} + \underbrace{(C(\cdot,t),C(\cdot,t))_H}_{C(t,t)} = d(t_n,t)^2 \to 0\,.$$ But by definition of $$H(C)$$ we have $$H(C) = \overline{\text{span}}\,\Phi(T)$$ and since $$T$$ is separable and $$\Phi$$ is continuous, the set $$\Phi(T)$$ is separable and hence also $$H(C)$$.
One can show that actually $$H(C)$$ is really a space of functions, i.e. it can be made to consist of true functions $$T \to \mathbb{R}$$ and not only equivalence classes of sequences.