I'm trying to understand this answer, since I'm stuck on the same proof as OP.
$2)$ A two-dimensional surface admits an almost complex structure if and only if it is orientable. If $X$ is an oriented surface, then choosing a Riemannian metric on $X$, we obtain the corresponding Hodge dual $\ast : T^*X \to T^*X$ which satisfies $\ast^2 = -\operatorname{id}_{TX}$. The Riemannian metric also gives rise to an isomorphism $\Phi_g : TX \to T^*X$ which allows us to construct the map $J : TX \to TX$ given by $J = \Phi_g^{-1}\circ\ast\circ\Phi_g$. As $J$ satisfies $J^2 = -\operatorname{id}_{TX}$, it is an almost complex structure on $X$.
As I have never worked with the Hodge operator before, could some one clarify to me how one would check why (if?) the thus obtained almost complex structure $J$ is metric-compatible?