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.

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" would be utterly absurd. The complete proof of the uniformization theorem is one of the hardest mathematical achievements of the early 20th century; the proof 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.