Redefining the Concept of "Path"
I'll answer the question under the assumption that the definition of path includes the requirement that it self-intersects at most a finite number of times. As I pointed out in this comment, I believe that in the context of Bruce Palka's textbook this assumption does not detract from the generality of my answer.
Some Preliminary Notations
For every path $\gamma : [a,b]\rightarrow\mathbb{C}$ ($a,b\in\mathbb{R}$, $a < b$) and every $c \in \mathbb{C}$ define
$$
\begin{align*}
\Xi(\gamma) &:= \big\{x\in[a,b] : \exists y\in[a,b]\setminus\{x\}.\ \gamma(x)=\gamma(y)\big\}\\
\xi(\gamma) &:= \big|\Xi(\gamma)\big|\\
\Omega(\gamma,c) &:= \big\{x\in[a,b] : \gamma(x)=c\big\}\\
\omega(\gamma,c) &:= \big|\Omega(\gamma,c)\big|
\end{align*}
$$
With these notations, we can state our assumption on paths formally: for every path $\gamma$, $\Xi(\gamma)$ is finite, or equivalently $\xi(\gamma) \in \{0,1,2,\dots\}$.
Note that $1\notin\operatorname{Rng}\xi$, and that for every closed path $\gamma:[a,b]\rightarrow\mathbb{C}$, $\xi(\gamma)\geq 2$, since $a, b \in \Xi(\gamma)$. A closed path $\gamma:[a,b]\rightarrow\mathbb{C}$ is simple iff $\xi(\gamma) = 2$.
A Restatement of the Problem
Let $D$ be a domain in $\mathbb{C}$, and let $z_0 \in \mathbb{C}\setminus D$.
It suffices to show that for every piecewise smooth path $\gamma$ in $D$ such that $n(\gamma,z_0)\neq0$, there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$.
In fact, if we show this, then by Lemma 2.1 (ii) and (iii) on p. 157 of Palka's textbook, $n(\gamma_1,z_0)\in\{-1,1\}$. If $n(\gamma_1,z_0)=1$, we're done. If $n(\gamma_1,z_0) = -1$, then $-\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(-\gamma_1,z_0) = -n(\gamma_1,z_0) = 1$, as desired.
Proof
Denote by $M$ the set of those $m \in \{2,3,\dots\}$ satisfying that for all closed, piecewise smooth paths $\gamma$ in $D$ for which $n(\gamma,z_0)\neq0$ and $\xi(\gamma) = m$, there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. We will now prove the claim if we show that $\{2,3,\dots\}\subseteq M$. We will do so by complete induction.
Base case: Let $\gamma$ be closed, piecewise smooth path in $D$ such that $n(\gamma,z_0)\neq0$ and $\xi(\gamma) = 2$. Set $\gamma_1 := \gamma$. Then $\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $2\in M$.
Inductive case: Assume that for some $m\in\{2,3,\dots\}$ it is the case that $\{2,3,\dots,m\}\subseteq M$, and let $\gamma:[a,b]\rightarrow\mathbb{C}$ be a closed, piecewise smooth path in $D$ such that $n(\gamma,z_0) \neq 0$ and such that $\xi(\gamma) = m+1$.
If $\omega\big(\gamma,\gamma(a)\big) > 2$, define
$$
\begin{align*}
c &:= \max\Big(\Omega\big(\gamma,\gamma(a)\big)\setminus\{b\}\Big)\\
\alpha &:= \gamma\big|_{[a,c]}\\
\beta &:= \gamma\big|_{[c,b]}
\end{align*}
$$
Then $\gamma = \alpha+\beta$, and both $\alpha$ and $\beta$ are closed, piecewise smooth paths in $D$. Then $n(\gamma,z_0) = n(\alpha,z_0) + n(\beta,z_0)$. Since by assumption $n(\gamma,z_0)\neq0$, either $n(\alpha,z_0)\neq0$ or $n(\beta,z_0)\neq0$. Say w.l.g. that $n(\alpha,z_0)\neq0$. Since $\xi(\alpha) < \xi(\gamma)$, by the induction hypothesis there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $m+1\in M$.
So assume henceforth that $\omega\big(\gamma,\gamma(a)\big) = 2$.
If $\gamma$ is simple, set $\gamma_1 := \gamma$. Then $\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $m+1\in M$.
So assume henceforth that $\gamma$ is not simple.
Then there is a finite sequence $x_0, x_1, \dots, x_k \in [a,b]$, for some odd $k \in \{3,4,\dots\}$, such that $a=x_0 < x_1 < \cdots < x_k=b$, and such that the following conditions are satisfied with $\alpha_i := \gamma\big|_{[x_{i-1},x_i]}$, $i \in \{1,2,\dots,k\}$:
- For every $i \in \{1,2,\dots,k\}$, $\alpha_i$ is a piecewise smooth path in $D$.
- $\gamma = \alpha_1 + \alpha_2 + \cdots + \alpha_k$.
- For every even $i \in \{2,3,\dots,k\}$, $\alpha_i$ is closed, and $\xi(\alpha_i) \leq m$.
- For every odd $i \in \{1,2,\dots,k-2\}$, $\alpha_i(x_i) = \alpha_{i+2}(x_{i+1})$, and $\alpha_1 + \alpha_3 + \cdots + \alpha_k$ is simple and closed and $\xi(\alpha_1 + \alpha_3 + \cdots + \alpha_k) \leq m$.
(I leave the verification of these facts to the reader. The fact that $\alpha_1 + \alpha_3 + \cdots + \alpha_k$ is simple will not be used below.)
Define $\ell := \frac{k-1}{2}$, and
$$
\begin{align*}
\beta_0 &:= \alpha_1 + \alpha_3 + \cdots + \alpha_k\\
\beta_1 &:= \alpha_2\\
\beta_2 &:= \alpha_4\\
&\vdots\\
\beta_{\ell} &:= \alpha_{k-1}
\end{align*}
$$
Then
- For every $i \in \{0,1,\dots,\ell\}$, $\beta_i$ is a closed, piecewise smooth path in $D$ with $\xi(\beta_i) \leq m$.
- $n(\gamma,z_0) = n(\beta_0,z_0) + \cdots n(\beta_{\ell},z_0)$.
(I leave the verification of these facts to the reader.)
Since by assumption $n(\gamma,z_0)\neq0$, there is some $i\in\{0,1,\dots,\ell\}$ such that $n(\beta_i,z_0)\neq0$. Then by the induction hypothesis there exists some simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. We conclude that $m+1 \in M$.
$\square$
