Let $\gamma: [a,b] \rightarrow \mathbb{C}$ be continuous and piecewise continuously differentiable but not necessarily closed.
I want to show that $$w \mapsto \frac{1}{2i\pi} \int_{\gamma}\frac{1}{z-w}\mathrm{d}z$$ from $\mathbb{C} \setminus \gamma([a,b]) \to \mathbb{C}$ is continuous.
$f(w)=\frac{1}{2i\pi} \int_{\gamma}\frac{1}{z-w}\mathrm{d}z=\frac{1}{2i\pi} \int_{a}^b\frac{1}{\gamma(t)-w}\gamma'(t)\mathrm{d}t=\frac{1}{2\pi i}\sum_{i=1}^n\int_{\xi_{i-1}}^{\xi_i}\frac{\gamma'(t)}{\gamma(t)-w}\mathrm{d}t$
$a=\xi_1<...<\xi_n=b.$
Now, because $\gamma$ is piecewise continuously differentiable for every $i=1...n$ $\gamma'(t)$ is continuous on $[\xi_{i-1},\xi_i]$ and because $\mathbb{C} \setminus \gamma([a,b])$ the denominator is also continuous therefore $f(w)$ is continuous.
Is that correct? Maybe there is a way arguing using parameter integrals
As you mentioned, we can invoke Continuity under Integral Sign theorem.
For this, you have to prove that $$(t,w) \mapsto \frac{\gamma'(t)}{\gamma(t)-w}$$ can be bounded by an integrable function depending only of $t$ in a neighborhood of $w$.
As $\gamma([a,b])$ is compact and $w \notin \gamma([a,b])$, you have $d=d(w, \gamma([a,b]) > 0$.
Hence $\vert \gamma(t) - z \vert >d/2 >0$ for $z \in B(w, d/2)$ and $$\left\vert \frac{\gamma'(t)}{\gamma(t)-z} \right\vert \le \frac{2 \sup\limits_{t \in [\xi_{i-1}, \xi_i]} \vert \gamma^\prime(t) \vert}{d}$$
for $(t, z) \in [\xi_{i-1}, \xi_i] \times B(w,d/2)$. And the RHS of inequality above is a constant.