If the curve $X$ (over the number field $k$) has no $k$-points at all, then Mordell's conjecture is true for $X$.Otherwise, if $O$ is a given $k$-point on $X$, we can get a map (the Albanese map)$X \to Jac(X)$ via $P \mapsto P - O$, embedding $X$ as a $k$-subvariety of $Jac(X)$.
The Mordell--Weil theorem shows that $Jac(X)(k)$ is a finitely generated abelian group.Since $g \geq 2$ by assumption, the curve $X$ is of positive codimension in $Jac(X)$,and so it is not unreasonable to imagine that $X$ intersects $Jac(X)(k)$ in only finitelymany points. (George Lowther's answer, posted while I was writing this, describes the sameheuristic.)
In fact, my understanding is that Weil proved the general form of the Mordell--Weil theorem(in his thesis, I think) precisely to try to implement this strategy and so prove Mordell'sconjecture. Unfortunately, no-one has been able to implement this strategy quite as cleanly as the intuition above suggests, although it inspired an important approach, due to Chabauty, which established the Mordell conjecture in some non-trivial cases; see this paper by McCallum and Poonen for an explanation of Chabauty's method, and Coleman's strengthening of it.
More recently, Minhyong Kim has developed an anabelian strengthening of Chabauty's method; see e.g. here and here. The second paper (joint with Coates) gives a new proof of Mordell for Fermat curve's, among other examples.
