Take a genus $>1$ curve $X_0$ with a rational point, equipped with a finite map to an elliptic curve $X_0\rightarrow E$, which has a point of order $n$. Let $X_i := X_{i-1}\times_{E} E$ be the base change by the multiplication by $n$ map $[n] : E\rightarrow E$. Since $[n]$ is etale (pick $n$ coprime to the characteristic of $k$), each $X_i$ is a smooth projective curve of genus $>1$, and by the universal property of fiber products, $|X_i(K)| = |E[n](K)|\cdot|X_{i-1}(K)|$.

EDIT: As Jason Starr mentioned, I should probably also arrange that each $X_i$ is connected (even though the OP didn't explicitly specify this!). This will be true if $X_0$ is a $G$-Galois branched cover of $E$ where $G$ is a finite perfect group.

Specifically, to see why this suffices, let $G$ be a finite 2-generated perfect group, ie a finite group generated by 2 elements such that $G = [G,G]$ (for example, any finite nonabelian simple group). Let $E^\circ$ denote a punctured elliptic curve (puncturing at the identity), then its etale fundamental group is isomorphic to $\widehat{F_2}$, the profinite free group on 2 generators. Let $p : \widehat{F_2}\rightarrow G$ be a surjection, which corresponds to a $G$-galois cover over $E^\circ$, which upon normalizing one obtains a ramified $G$-galois cover over $E$ which is a connected smooth projective curve of genus $>1$ (genus $>1$ because the map is ramified, ramified because $G$ is nonabelian). Take this curve to be $X_0$. I claim that the sequence $X_i := X_{i-1}\times_E E$, where the base change is via the map
$$[n^i] : E\rightarrow E$$
is a sequence which gives a counterexample to your questions. By the above discussion each $X_i\rightarrow X_{i-1}$ is a finite etale map of smooth projective curves of genus $>1$. It remains to show that each $X_i$ is connected. This is true because $G$ is generated by commutators (ie, is perfect).

In more detail, the map $[n^i] : E\rightarrow E$ restricted to the preimage of $E^\circ$ gives a finite etale cover over $E^\circ$, corresponding to a subgroup $P_i\le\widehat{F_2}$ of index $n^{2i}$. Specifically
$$P_i = \text{Ker}\left(\widehat{F_2}\rightarrow(\mathbb{Z}/n^i\mathbb{Z})^2\right)$$
and is by Galois theory the fundamental group of $[n^i]^{-1}(E^\circ)$.
Hence, $P_i$ contains the commutator subgroup $[\widehat{F_2},\widehat{F_2}]\le\widehat{F_2} = \pi_1(E^\circ)$. Then, by Galois theory, the restriction of the pulled back $G$-Galois cover $X_i\rightarrow E$ to $E^\circ$ corresponds to the composition
$$P_i\hookrightarrow\widehat{F_2}\stackrel{p}{\rightarrow} G$$
which is surjective since $G$ is perfect, and $P_i$ contains $[\widehat{F_2},\widehat{F_2}]$. By the Galois correspondence, this implies that $X_i$ is connected.