Some comments alluded to the possibility to show this using the Riemann uniformization theorem (by paracompactness, any oriented $2$-manifold has an almost complex structure, which is integrable by Newlander-Nirenberg and by the uniformization theorem, it will be biholomorphically equivalent to the plane or the unit disc, hence diffeomorphic to $R^2$). This is not circular, but to claim that this is "simple" utterly absurd; the proof of the RUT uses a lot of analysis and also a bit of algebraic topology.
Using Morse theory, you can argue as follows: let $U$ be a connected noncompact surface, pick a Morse function $f: U \to \mathbb{R}$. One can modify $f$ so that it has no critical points of index $2$ and precisely one critical point of index $0$, so let us assume that $f$ has this property. This is the most basic case of the handle-cancellation technique.
Now let $C_{\ast}(f)$ be the chain complex of the Morse function. $C_k (f)$ has the critical points of index $k$ as a basis. If $f$ is above, it follows that $C_0 (f)=Z$, $C_k (f)=0$ if $k \geq 2$. The differential $C_1 \to C_0$ will be zero and so $H_1 (U)= C_1 (f)$.
If $H_1 (U)=0$, we see that there is a Morse function $f:U \to \mathbb{R}$ with precisely one minimum. Use the flow lines of $f$ to cook up a diffeomorphism $f: U \to \mathbb{R}^2$.
I don't think you get this result much cheaper.