The question in the title naively appears to be true, for in a compact metric $X$ space every sequence has a convergent subsequence. If $(a_n)$ traverses a compact curve $\Gamma$ then there is a convergent subsequence $(b_n)$. But since (a_n) and (b_n) both traverse the same curve, they should converge to the same limit.
But this is very counterintuitive. For example, I would not expect a sequence along some of the wilder space filling curves to converge, so I doubt my intuition.
EDIT: With a curve $\Gamma$ parametrized by a continuous $\gamma: [0,1] \to X$
Your intuition may come from the fact that you're considering a compact curve in $X$ as being a continuous map $f:[0,1]\to X$ and a "sequence along a compact curve" as a sequence $u_n=f(x_n)$ where $x_n$ is an increasing sequence of points in $[0,1]$. In this case, your intuition is correct, since the sequence $x_n$ must converge to some limit $\ell$ (being an increasing and bounded sequence of reals) and therefore by continuity $u_n$ will converge to $f(\ell)$.