Assume $X$ is a metrix space. In the case that $X$ is an inverse limit of one-dimensional finite CW complexes, it is known that $K_0(C(X))$ and $K_1(C(X))$ are torsion free . So I tried some examples where $X$ are two-dimensional finite CW complex, but faild to construct one that has non-torsion-free $K$-groups.
For instance, I considered the case that $X$ is constructed by attaching a 2-dimensional disc $D^2$ to $S^1$ via a degree 2 map. Then there is a short exact sequence
$$C_0(\mathbb R^2)\to C(X)\to C(S^1)$$ which gives a six-term exact sequence of $K$-groups
$$\require{AMScd} \begin{CD} \mathbb Z @>>> K_0(C(X)) @>>> \mathbb Z\\ @AAA @. @VVV \\ \mathbb Z @<<< K_1(C(X)) @<<< 0 \end{CD}$$
Since the boundary map on the left side is an isomorphism ($K_1(S^1)\simeq K_1(C_0(\mathbb R))\simeq K_0(C_0(\mathbb R^2))$), $K_*(C(X))$ is torsion-free.
I also tried to attach more components to $X$, but none could change the fact that the boundary maps are too simple.
Is there an example that $C(X)$ has non-torsion-free $K$-groups? I really can not imagine when this could happen.
No definitely not.
Every abelian group appears as the $K_0$-group or the $K_1$-group of a commutative $C^*$-algebra. This is proven in Blackadar's K-theory book.