Let $\gamma$ be a (not necessarily closed) curve in $\mathbb{C}$, and $\phi: \text{Im}(\gamma) \to \mathbb{C}$ be a continuous function. I want to show that
$$f(z)=\int_{\gamma}\frac{\phi(\lambda)}{\lambda-z}d\lambda$$
is holomorphic on $\mathbb{C}\backslash \text{Im}(\gamma)$.
Here is my attempt at a solution, I want to make sure that this is correct since it uses many different theorems from complex analysis, and I need to make sure all their conditions are satisfied.
Note $\mathbb{C}\backslash \text{Im}(\gamma)$ is open in $\mathbb{C}$, so if for every piecewise $C^1$ curve $\zeta$ on $\mathbb{C}\backslash \text{Im}(\gamma)$
$$\oint_{\zeta}f(z)dz=0,$$
then it is holomorphic by Morera’s theorem.
\begin{align*} \oint_{\zeta}f(z)dz&= \oint_{\zeta} \int_{\gamma} \frac{\phi(\lambda)}{\lambda-z}d\lambda dz \\ &=\int_{\gamma} \oint_{\zeta} \frac{\phi(\lambda)}{\lambda-z}dzd\lambda \quad \text{by Fubini’s theorem} \end{align*}
Since $\phi$ is continuous, and $\lambda \neq z$ since $z \in \mathbb{C}\backslash \text{Im}(\gamma)$, $\lambda \in \text{Im}(\gamma)$, $\frac{\phi(\lambda)}{\lambda-z}$ is holomorphic on $\zeta$. $\mathbb{C}\backslash \text{Im}(\gamma)$ is a simply connected domain if $\gamma$ is not closed. If $\gamma$ is closed, the domain is split into two simply connected parts, and $\zeta$ must lie on only one of those parts since we have established it does not cross $\gamma$. These conditions are sufficient for Cauchy’s integral theorem, which tells us
$$\oint_{\zeta} \frac{\phi(\lambda)}{\lambda-z}dz =0,$$
yielding the result upon application of Morera’s theorem.
Is this argument correct?
As already said in the comments, $\mathbb{C}\setminus\operatorname{Im}(\gamma)$ may not be simply connected.
But holomorphy is a local property: It suffices to show that $f$ is holomorphic in every open disk $D \subset \mathbb{C}\setminus\operatorname{Im}(\gamma)$. And then your proof works because you only need to consider closed curves $\zeta$ in the (simply-connected) disk $D$, where $$ \tag{*} \oint_{\zeta} \frac{\phi(\lambda)}{\lambda-z}\,dz = \phi(\lambda)\oint_{\zeta} \frac{dz}{\lambda-z} = 0 $$ by Cauchy's integral theorem.
The continuity of $\phi$ is not needed in $(*)$, but when applying Fubini's theorem. This condition can be relaxed to requiring that $\int_\gamma |\phi(\lambda)| \, d\lambda$ exists.