I guess you need the equicontinuity for $f$ in the unit ball of $\mathscr{C}^0([0,1])$, and not the whole space (which cannot be, check it !). For this, first notice that given $f$ in the unit ball of $\mathscr{C}^0([0,1])$, and $x,z\in[0,1]$ \begin{align*} |Jf(x)-Jf(z)| \leq \int_0^1 \left|\frac{1}{|x-y|^c}-\frac{1}{|z-y|^c}\right|\mathrm{d}y, \end{align*} so that we just have to check the continuity of $Jf$, where $f=1$. Now in this case, for $(x_n)_n\in[0,1]^\mathbb{N}$ such as $(x_n)_n\rightarrow a\in[0,1]$, we have by (two) change(s) of variable \begin{align*} Jf(x_n) = \int_0^1 \frac{1}{|x_n-y|^c} \mathrm{d} y = \int_{x_n-1}^{x_n} \frac{1}{|u|^c}\mathrm{d}u \operatorname*{\longrightarrow}_{n\rightarrow+\infty} \int_{a-1}^{a} \frac{1}{|u|^c}=Jf(a),\end{align*} by dominated convergence, since \begin{align*} \mathbf{1}_{[x_n-1,x_n]}(t)\frac{1}{|t|^c} \leq \mathbf{1}_{[-1,1]}(t)\frac{1}{|t|^c} \in L^1(\mathbb{R}), \end{align*} since $c<1$.