 Solution

The Arzela-Ascoli theorem says that a subset $\mathcal{F}\subseteq C(K)$ is relatively compact in the uniform norm exactly when it is equicontinuous and pointwise bounded.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subseteq C(K,\mathbb{R})$ is uniformly dense if it contains the constants and separates points of $K$. In the complex version, a complex subalgebra $A\subseteq C(K,\mathbb{C})$ must additionally be closed under complex conjugation.

For an example, take $K=[0,1]$ andThis family is bounded in the uniform norm, but it is not equicontinuous at $1$. Equivalently, every subsequence converges pointwise to the discontinuous function that is zero on $[0,1)$ and one at $1$, so no subsequence converges uniformly. Thus $\mathcal{F}$ is bounded but not relatively compact.