Take a metric on your line bundle and a metric connection. This connection cannot have curvature, and as the 2-sphere is simply connected, there is no monodromy, so parallel transport trivializes the bundle. Of course, this is only the differential geometry version of the answers of Francesco Polizzi and Dan Petersen.
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SebastianNov 20 '12 at 12:02

4 Answers
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Since $GL(1,\mathbf R)$ and the symmetric group $S_2$ are homotopy equivalent as topological groups, there is a bijective correspondence between isomorphism classes of real line bundles and double covers. It follows that a simply connected space has no nontrivial real line bundles.

In general, there is a bijection $$\Phi \colon [S^{k-1}, \textrm{GL}_n^+(\mathbb{R})] \to \textrm{Vect}_+^n(S^k),$$
where $[X,Y ]$ denotes the set of homotopy classes of continous maps from $X$ to $Y$ and $\textrm{Vect}_+^n(Z)$ denotes the set of isomorphism classes of real (oriented) vector bundles of rank $n$ on $Z$.

In the case of line bundles we have $n=1$ and $\textrm{SO}(1)$ is just a point, hence the bijection $\Psi$ shows that any orientable real line bundle on $S^k$ is trivial. Furthermore, it is not difficult to prove that when $k \geq 2$ any real vector bundle on $S^k$ is orientable, so any real line bundle is trivial.

Notice that this is not true for $S^1$, where there is an orientable bundle (the trivial one) and a nonorientable one (the Moebius band).

Similar methods of classification can be applied to the case of complex vector bundles in order to show that $S^2$ admits nontrivial complex line bundles.

In fact, one finds that there is a bijection $$\Phi_{\mathbb{C}} \colon [S^{k-1}, \textrm{GL}_n(\mathbb{C})] \to \textrm{Vect}_{\mathbb{C}}^n(S^k).$$
For $n=1$, moreover, there is a bijection between $[S^{k-1}, \textrm{GL}_1(\mathbb{C})]$
and $H^2(S^k, \mathbb{Z})$. In particular, if $k \neq 2$ the only complex line bundle on $S^k$ is the trivial one, whereas if $k =2$ there is a discrete family $\{L_t \}$ of non-isomorphic complex line bundles, parametrized by $t \in \mathbb{Z}$. In fact, viewing $S^2$ as the Riemann sphere with complex coordinate $z$, the transiction function of $L_t$ is $z^t$.

Another solution is to consider clutching functions. The sphere $S^2$ is obtained by gluing two disks along their boundaries, it is not hard to prove that the information is encoded by a continuous map:
$$f:S^1\rightarrow GL_1(\mathbb{R})$$
and that we have a map:
$$[S^1,GL_1(\mathbb{R})]\rightarrow Vect^1(S^2)$$
where on the left hand side you have homotopy classes of clutching functions and on the right hand side you have isomorphism classes of real line bundles over $S^2$. As our continuous map $f$ is homotopic to a constant map the associated vector bundle is trivial.
In general for $n$-dimensional real vector bundles over $S^k$ you will have a map:
$$[S^{k-1},GL_n(\mathbb{R})]\rightarrow Vect^n(S^k).$$
This map becomes a bijection when you consider oriented vector bundles.

edit: I just realize that I write my post at the same time as Francesco.

Along the same lines. Rank $1$ real vector bundles over a compact CW complex $X$ are all pullbacks of the tautological line bundle over $\mathbb{RP}^\infty$. The space of isomorphism classes of such bundles can be identified with the space $[X,\mathbb{RP}^\infty]$ of homotopy classes of continuous maps $X\to \mathbb{RP}^\infty$. Since $\mathbb{RP}^\infty$ is the Eilenberg-MacLane space $K(\mathbb{Z}/2, 1)$ (the universal cover $S^\infty$ is contractible) we deduce that $[X,\mathbb{RP}^\infty]$ can be identified with the cohomology group $H^1(X,\mathbb{Z}/2)$. When $X=S^2$ this group is trivial.