Can one disjoin any submanifold in $\mathbb R^n$ from itself by a $C^{\infty}$-small isotopy? Let $M$ be a manifold and $V$ be an oriented vector bundle. It's well known that if the Euler class of $V$ is non zero, then $V$ can't have a non-vanishing section. The converse is not true, see Vanishing of Euler class , or the answer by John Klein below, given to the first version of this question. So:
Question. Recall that the normal bundle of a smooth orieantable submanifold in $\mathbb R^n$ always has zero Euler class.  Does such a normal bundle always has a non-vanishing section? If yes, how to prove this? If no, what is a counterexample? 
PS. It turns out that this question is a subquestion of the following much more informed one: 
Embeddings without nonvanishing normal vector fields
 A: Here's a counter example. Take any embedding of $\mathbb{C} P^2$ in $\mathbb{R}^7$ (such embeddings exist by 
Steer, B., On the embedding of projective spaces in Euclidean space, Proc. Lond. Math. Soc., III. Ser. 21, 489-501 (1970). ZBL0206.25501;
the construction is summarised in this answer on MSE).
If such an embedding admits a normal vector field, then by the Compression Theorem of Rourke and Sanderson it is isotopic to an embedding with normal field parallel to the last coordinate of $\mathbb{R}^7=\mathbb{R}^6\times\mathbb{R}$, and then the projection is an immersion $\mathbb{C} P^2\looparrowright \mathbb{R}^6$. Such immersions cannot exist, by András Szűcs' answer here.
A: There are lots of counter-examples. 
Here's one:
The fibration $\text{SO}(3) \to \text{SO}(4) \to S^3$ splits, since it is the principal bundle of the tangent bundle of $S^3$, and the latter is parallelizable. In particular,  $\pi_5(\text{SO}(4)) \cong \pi_5(\text{SO}(3)) \oplus \pi_5(S^3) \cong \Bbb Z/2 \oplus \Bbb Z/2$ . 
Let $a$ and $b$ denote the generators for the summands. Then $(a,b)$ determines a rank $4$-vector bundle over $S^6$ with trivial Euler class (since $H^4(S^6) = 0$). But 
this vector bundle does not have a section, for if it did that would lead to the contradiction that $b=0$.
