Let $f:[c,d] \rightarrow \mathbb{R}$ be a continuous function which is piecewise analytic, i.e., there exists $N\in \mathbb{N}$ and an increasing sequence $\left(i_{n}\right)_{1 \le n \le N}$, $i_{1}=c, i_{N}=d,$ such that $$ \forall x \in(c, d) \quad f(x)=\sum_{n=1}^{N-1} \chi_{\left[i_n, i_{n+1}\right)}(x) f_{n}(x), $$ where $f_n$ are analytic functions defined on $\left[i_{n}, i_{n+1}\right]$ and $\chi_{\left[i_{n}, i_{n+1}\right)}$ are characteristic functions for $n=1, \ldots, N-1$.
Consider the set $Z(f)$ of zeros of $f$ in $(c,d)$.
Does the set $Z(f)$ have a finite number of connected components?
If there are infinitely many components of $Z(f)$, then there must be an $1\le n\le N$ such that $(i_n,i_{n+1})$ contains infinitely many of them, which we can take to be countable, say $(C_j)_{j\in \mathbb N}$ Now, choose $x_j\in C_j$. Then the sequence $(x_j)_{j\in \mathbb N}$ has a limit point $x\in [i_n,i_{n+1}]$ but $x\neq (i_n,i_{n+1})$ for otherwise $f=0$ on $(i_n,i_{n+1})$ (because $f_n$ is analytic), which is a contradiction since $(i_n,i_{n+1})$ is connected. Without loss of generality, $f(i_n)=0.$ But now, note that $f_{n-1}\cup f_n$ is analytic on $(i_{n-1},i_{n+1})$ and $x$ is a limit point of a sequence in $Z(f)$ interior to this interval, so again we have a contradiction because, in this case, $f=0$ on $(i_{n-1},i_{n+1}).$