= Solution
The <Arzelà-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 $K=[0,1]$ and
$$
\mathcal F=\{x\mapsto x^n:n\geq1\}.
$$
This 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 \b[$\mathcal F$ is bounded but not relatively compact.]
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