If $X$ is a paracompact topological space, $\operatorname{Vect}_k(X) \cong [X, BO(k)] = [X, \operatorname{Gr}_k(\mathbb{R}^{\infty})]$ where $\operatorname{Vect}_k(X)$ denotes isomorphism classes of real rank $k$ vector bundles on $X$ and the square brackets denote homotopy classes of maps. In particular,
$$\operatorname{Vect}_1(X) \cong [X, \operatorname{Gr}_1(\mathbb{R}^{\infty})] = [X, \mathbb{RP}^{\infty}] = [X, K(\mathbb{Z}_2, 1)] \cong H^1(X; \mathbb{Z}_2).$$
Under this identification, the isomorphism class of a line bundle $L$ on $X$ is identified with its first Stiefel-Whitney class $w_1(L)$. So $L$ is trivial if and only if $w_1(L) = 0$ (i.e. $L$ is orientable).
Every open subset of $\mathbb{R}^n$ is paracompact (metric spaces are paracompact), so we can use the above characterisation in order to answer your question.
The open set $U = \mathbb{R}^2\setminus\{(0,0)\}$ in $\mathbb{R}^2$ deformation retracts onto $S^1$, so
$$\operatorname{Vect}_1(U) \cong H^1(U; \mathbb{Z}_2) \cong H^1(S^1; \mathbb{Z}_2) \cong \mathbb{Z}_2.$$
Therefore, there is a non-trivial line bundle on $U$.
However, if we take $U$ to be contractible, then the claim holds.
Note, there are non-contractible open sets of $\mathbb{R}^n$ such that every vector bundle on them is trivial. For example, $U = \mathbb{R}^4\setminus\{(0,0,0,0)\}$ which deformation retracts onto $S^3$. Either by the above characterisation or by using the clutching construction, we see that $\operatorname{Vect}_k(U) \cong \pi_2(O(k))$. As the second homotopy group of any Lie group is zero, we see that every vector bundle on $U$ is trivial.