The Volterra operator is given as \begin{eqnarray} (Vf)(x)=\int_0^xK(x,y)f(y)\,{\rm d}y. \end{eqnarray}
By the Arzelà–Ascoli theorem, $V\colon C^0[0,1]\rightarrow C^0[0,1]$ is compact operator. But, if $V\colon C^0[0,1]\rightarrow C^1[0,1]$, is this a compact operator?
Rule of thumb: compactness should be expected when $Tf$ is at least a little bit nicer than a generic element of the target space. Integration against a continuous kernel does this because $Tf$ inherits modulus of continuity from the kernel, which is an improvement compared to having totally unknown modulus of continuity.
But for compactness from $C^0$ to $C^1$, you need a lot more improvement: $Tf$ now must be nicer than a generic $C^1$ function. Even with $K\equiv 1$ you don't get this: the range of $T$ in this case consists of all $C^1$ functions vanishing at $0$. And since the $C^1$ norm of $Tf$ includes the supremum of derivative, $T $ is actually norm-preserving in this case, hence not compact.