The holonomy group need not be compact. For example, take $S^1$, trivialize its tangent bundle and let $\Gamma_{1,1}^1 = 1$, constant on $S^1$. If you parallel transport any vector around $S^1$, the holonomy is multiplication by a non-zero number.

So it's a discrete holonomy, but still countably-infinite.

The holonomy group need not be compact. For example, take $S^1$, trivialize its tangent bundle and let $\Gamma_{1,1}^1 = 1$, constant on $S^1$. If you parallel transport any vector around $S^1$, the holonomy is multiplication by some number, call it $k$, $k > 0$ and $k \neq 1$. If we use the standard counter-clockwise and euclidean unit vector trivialization of $TS^1$, I suppose $k = e^{-2\pi}$. So the holonomy $\pi_1 S^1 \to Hom(T_1 S^1)$ is the map that sends $n \in \mathbb Z \equiv \pi_1 S^1$ to multiplication by $e^{-2\pi n}$ in $T_1 S^1$.

So it's a discrete holonomy, but still countably-infinite.