Theorem: If $E$ is a separable metric space and $K$ is continuous on $E \times E$, $\mathbb{H}(K)$ is separable and the functions in $\mathbb{H}(K)$ are continuous on $E$.
The notation $\mathbb{H}(K)$ denotes the RKHS with reproducing kernel $K$.
Proof: By assumption, there is a countable set $\{t_1, t_2, \ldots\}$ that is dense in $E$. If $f \in \mathbb{H}(K)$, $$ |f(t)-f(s)|=|\langle f, K(\cdot, t)-K(\cdot, s)\rangle| \leq\|f\|\|K(\cdot, t)-K(\cdot, s)\| . $$
However, $$ \|K(\cdot, t)-K(\cdot, s)\|^2=K(t, t)-2 K(t, s)+K(s, s) $$ which converges to zero as $t \rightarrow s$ by the continuity of $K$.
The collection of functions $$ \{\sum_{i=1}^n a_i K\left(\cdot, t_i\right): a_i \in \mathbb{Q}, n \in \mathbb{Z}^{+}\, $$ with $\mathbb{Q}$ denoting the set of rationals, is dense in $\mathbb{H}(K)$. This implies that $\mathbb{H}(K)$ is separable.
Question: If $f \in \mathbb{H}(K)$, $|f(t)-f(s)|$ converges to zero as $t \rightarrow s$. It looks like the functions in $\mathbb{H}(K)$ are not only continuous on $E$, but they are also uniformly continuous on $E$. Am I correct?
Specifically, for any $\epsilon > 0$, we can choose $\delta > 0$ such that $|K(\cdot, t) - K(\cdot, s)| < \epsilon / |f|$ whenever $d(t, s) < \delta$, where $d$ is the metric on $E$. Then, for any $t, s \in E$ with $d(t, s) < \delta$, we have:
$$|f(t) - f(s)| \leq |f| |K(\cdot, t) - K(\cdot, s)| < |f| \cdot \frac{\epsilon}{|f|} = \epsilon$$
This shows that $f$ is uniformly continuous on $E$, as the $\delta$ that makes $|f(t) - f(s)| < \epsilon$ does not depend on the specific points $t$ and $s$, but only on $\epsilon$ and the function $f$.