Let $(X,d)$ be a compact metric space and $(x_n)$ a sequence in $X$ such that $d(x_n, x_{n+1}) \to 0$ when $n \to \infty$.
The set $\Gamma$ of the adherent values of $(x_n)$ is connected. Here is a proof of this. By adherent value I mean that $x\in X$ is an adherent value if all neighbourhood of $x$ contains infinitely many terms of $(x_n)$, that is $$ \Gamma = \bigcap_{p\in \mathbb N} \overline{\{x_n : \ n \ge p\}}.$$
Let $A$ and $B$ two disjoint non-empty closed subsets of $\Gamma$ such that $\Gamma = A\cup B$. $A$ and $B$ are compact as closed subsets of $X$ which is compact, hence $\alpha= d(A,B) >0$. Let $$A'= \{ x\in X : d(x,A) <\alpha/3 \}, \quad B'= \{x\in X : d(x,B) <\alpha/3\} \quad \text{and} \quad K = (A'\cup B')^c.$$
Then $x_n \in A'$ for infinitely many $n$, and the same holds for $B'$.
Let $N\in \mathbb N$ such that
$$\forall n \ge N, \quad d(x_{n},x_{n+1}) < \alpha/3. $$
There exists $N_a \ge N$ such that $x_{N_a}\in A'$, and $N_b >N_a$ such that $x_{N_b} \in B'$. There exists $N_a \le k_0 \le N_b$ such that $x_{k_0}\in K$, othewise there would be a $N_a \le k \le N_b$ such that $x_{k}\in A'$ and $x_{k+1}\in B'$, hence $d(x_{k},x_{k+1}) \ge \alpha/3$, which is false because $k \ge N$.
This way we build $k_0<k_1 < k_2 < \dots$, such that $x_{k_j} \in K$ for all $j\in \mathbb N$. $K$ is compact (closed in $X$), therefore $(x_{k_j})$ has an adherent value $l\in K$, which is also an adherent value of $(x_n)$ that is not in $\Gamma$, contradiction.
I'd like to know if the set $\Gamma$ could somehow not be pathwise-connected ?
If we could find a sequence of positive real numbers $(t_n)$ that satisfies $t_n-t_{n+1} \to 0$, such that $[0,1]$ is the set of adherent values of $(t_n)$, then the sequence $u_n = (t_n, \sin(1/t_n))\in \mathbb R^2$ would have the topologist's sine curve as its set of adherent points. I think of something like $$t_n = \frac{1+\cos(1/n)}{2}, $$ but I don't know if this works.
In fact, there are no restrictions on $\Gamma$ besides connectedness and compactness. That is, if $X$ is a compact metric space and $A\subseteq X$ is a closed connected subset, then there is a sequence $(x_n)$ in $X$ whose set of adherent values is $A$ such that $d(x_n,x_{n+1})\to 0$.
First, note that we may as well assume that $X=A$ (just replace $X$ with the subspace $A$). Fix a countable dense subset $\{y_k\}_{k\in \mathbb{N}}\subseteq X$. Note that since $X$ is connected, for any $\epsilon>0$ and any $a,b\in X$, there exists a finite sequence $a=c_1,c_2\dots,c_m=b\in X$ with $d(c_i,c_{i+1})<\epsilon$ (proof: writing $a\sim b$ if such a sequence exists, $\sim$ is an equivalence relation on $X$ whose equivalence classes are open and hence clopen). Now take a sequence $(x_n)$ that starts at $y_0$, then moves to $y_1$ by a finite sequence of steps where the distance is at most $1$, then moves to $y_2$ by a finite sequence of steps where the distance is at most $1/2$, and so on. This sequence will satisfy $d(x_n,x_{n+1})\to\infty$ and every point of $X$ will be an adherent point since $\{y_k\}$ is dense in $X$ (note that by connectedness $X$ has no isolated points unless it has only one point).
More generally, this argument works for any separable metric space and a closed subset $A\subseteq X$ which is "$\epsilon$-connected" for all $\epsilon>0$ in the sense used above (any two points can be linked by a finite sequence with distances $<\epsilon$ in each step). For compact spaces being $\epsilon$-connected for all $\epsilon$ is equivalent to connectedness, but in general it is weaker (for instance, $\mathbb{R}\setminus\{0\}$ is $\epsilon$-connected for all $\epsilon$).