A nowhere vanishing vector field $X$ tangent to the leaves of the foliation? I want to know if the following statement is true or false:
For every manifold $M$ and for every foliation on $M$ whose leaves have dimension $1$, there exists a nowhere vanishing vector field $X$ on $M$ tangent to the leaves of the foliation.
I think this is false. I was trying to come up with a counter example, namely the Klein Bottle. See here for a construction of the Klein Bottle.
I would say: draw a vector field vertically downwards (from top to bottom) on the square (parallel to the red lines). Then after construction, this would be a nowhere vanishing vector field on the Klein Bottle. A foliation whose leaves are 1-dimensional would be circles, see e.g. here
I'm not sure if this vector field would be tangent to each circle or not?  
 A: Given a foliated manifold $M$, we have a splitting $TM \cong T\mathcal F \oplus N \mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable). 
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T \mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $\eta$ is a real line bundle on $Y$, $E = \eta \oplus \eta \oplus \Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(\eta)^2$. So if the cup-square map $H^1(Y;\Bbb Z/2) \to H^2(Y;\Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $\eta$ is a summand of $TY$ as desired, and non-orientable foliations exist. 
In particular, because $H^1(M;\Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;\Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
