This is a theorem of Lang's from 1956. Here's an online document giving a proof (in the form $H^1(A,k)=0$):
Lecture 14: Galois Cohomology of Abelian Varieties over Finite Fields, William Stein. http://wstein.org/edu/2010/582e/lectures/582e-2010-02-12/582e-2010-02-12.pdf
Stein notes that there is a "more modern proof" in the first few sections of Chapter VI of Serre's Algebraic Groups and Class Fields.
The original article is Serge Lang, "Abelian varieties over finite fields," Proceedings of the National Academy of Sciences 41.3 (1955): 174-176. It is available at http://www.pnas.org/content/41/3/174.short. It's very short, but phrased in the "old-style" Weil language of algebraic geometry.
Here's a quick sketch of a proof with $k=\mathbb F_q$. Choosing a point of $X(\overline{k})$, we can make $X$ into an abelian variety over $\overline{k}$. Then the $q$-power Frobenius map $\phi_q:X\to X$ is the composition of an isogeny and a translation, say $\phi_q(x)=f(x)+x_0$ with $f:X\to X$ an isogeny (defined over $\overline{k}$) and $x_0\in X(\overline{k})$. The fact that $\phi_q$ is purely inseparable inseparable implies that $f$ is purely inseparable, and hence $1-f$ is separable (look at the action of $(1-f)^*$ acts as the identity map on differentials). Use this to conclude thatThus $1-f$ has finite kernel, so it is surjective, and thus there is a point $x_1\in X(\overline k)$ satisfying $(1-f)(x_1)=x_0$. This implies that $\phi_q(x_1)=x_1$, and hence that $x_1\in X(k)$.