Proving a nowhere vanishing vector field on 2D manifold implies $TU\cong M\times S^1$

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So, I am trying to solve the following problem. Suppose you have a nowhere zero smooth vector field on a 2 dimensional oriented compact manifold. Prove that the unit tangent bundle $TU$ is diffeomorphic to $M\times S^1$.

So, I can see that at each point of $M$ the unit tangent space is isomorphic to $S^1$, but I don't see how to use the nowhere vanishing vector field.

First, I need to convince myself that over small enough neighborhoods we have a local trivialization. Then maybe by compactness I could cover the manifold with finitely many trivially covered neighborhoods. And maybe then use the nowhere vanishing vector field to connect them somehow, but I am not sure.

Or, maybe there is a clever way of writing the diffeomorphism using the vector field.

Any hints or thoughts would be appreciated, thanks in advance!

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First note that if $V$ is a nowhere vanishing vector field, then $\widehat{V} = \frac{V}{\|V\|}$ is a global section of the unit tangent bundle.

Let $p \in T_mU \cong S^1$, then there is $z \in S^1$ such that $p = z\widehat{V}(m)$. The diffeomorphism $TU \cong M\times S^1$ is given by $p \mapsto (\pi(p), z)$.

This is very similar to the proof that a line bundle which admits a nowhere zero section is trivial.