Folliation and non-vanishing vector field.

Solution 1:

I assume that your manifold and foliation are smooth and foliation is of codimension 1, otherwise see Jack Lee'a comment. Then pick a Riemannian metric on $X$ and at each point $x\in M$ take unit vector $u_x$ orthogonal to the leaf $F_x$ through $x$: There are two choices, but since your foliation is transversally orientable, you can make a consistent choice of $u_x$. Then $u$ is a nonvanishing vector field on $X$.

In fact, orientability is irrelevant: Clearly, it suffices to consider the case when $X$ is connected. Then you can pass to a 2-fold cover $\tilde{X}\to X$ so that the foliation ${\mathcal F}$ on $X$ lifts to a transversally oriented foliation on $\tilde{X}$. See Proposition 3.5.1 of

A. Candel, L. Conlon, "Foliations, I", Springer Verlag, 1999.

You should read this book (and, maybe, its sequel, "Foliations, II") if you want to learn more about foliations.

Then $\chi(\tilde{X})=0$. Thus, $\chi(X)=0$ too. Now, recall that a smooth compact connected manifold admits a nonvanishing vector field if and only if it has zero Euler characteristic. Thus, $X$ itself also admits a nonvanishing vector field.

Incidentally, Bill Thurston proved in 1976 (Annals of Mathematics) that the converse is also true: Zero Euler characteristic for a compact connected manifold implies existence of a smooth codimension 1 foliation. This converse is much harder.