Line subbundles of tangent bundle of a manifold is trivial? I was pretty sure about this result but don't know how to prove it. I will state the question again:
Is any smooth line subbundle (or equivalently smooth 1-dimensional distribution) of the tangent bundle of a manifold is always trivial? Namely, once you have a 1-dim distribution on a manifold, you can have a nowhere vanishing vector field on that.
How to prove?
Thank you in advance!
 A: This is not true. Consider the Klein bottle $K$.
As $K$ is a closed manifold with $\chi(K) = 0$, it admits a nowhere-zero vector field. The orthogonal complement of such a vector field is a line subbundle $L$ of $TK$. If $L$ were trivial, then $TK \cong L\oplus\varepsilon^1$ would be trivial, but this is impossible as $K$ is non-orientable.
However, as this example illustrates, if $TM$ admits a line subbundle, then it also admits a trivial line subbundle, i.e. $M$ admits a nowhere-zero vector field. See this answer.
A: The non-orientable foliation below on the punctured disk shows that a $1$-dimensional distribution does not generally have a continuous nowhere vanishing section.
That doesn't mean the punctured disk has no non-vanishing vector field, of course, it just means that we can't generally select a continuous, non-vanishing vector field from a given distribution.
The field of lines on a Möbius strip is another example, but the punctured disk may be more impressive since its tangent bundle is trivial.

