From the Classification Theorem for closed (i.e. compact and boundaryless) surfaces, it follows that $S^2$ is the only closed surface with trivial $\pi _1$. That's easy because the fundamental group classifies closed surfaces.
I'd like to get the same conclusion (i.e. that a simply-connected closed surface is isomorphic to $S^2$, in your favorite category) without that theorem. Is it possible? Any discussion about (generalized) Poincaré conjecture I could find starts saying that in dimension $1$ and $2$ it's true because of the classification theorems (available in those dimensions), dimension $3$ was settled by Perelman and then switches to high-dimensional wild cases. No insights on a direct proof in dimension $2$.
My attempt: let $S$ be a closed surface with $\pi_1(S)=1$; trivial $\pi_1$ implies that $S$ is orientable, that $S$ is covered only by itself, that every embedded loop bounds a disk (so $S$ has genus $0$, since cutting along every embedded loop disconnects $S$); compactness rules out $\mathbb{R}^2$ and $\partial S = \emptyset$ rules out the unit disk. But here I don't know how to go on. Maybe one should try to find an explicit isomorphism to $S^2$, but I can't see how.
Best Answer
The following proof is a bit over the top, but here it is:
Prove Gauss' isothermal coordinates theorem (every 2-dimensional Riemannian manifold is conformally flat). As a corollary, conclude that every oriented 2-dimensional Riemannian manifold $(M,g)$ has a natural structure of a Riemann surface (the holomorphic atlas is given by charts $\phi: U\subset {\mathbb C} \to V\subset M$ such that $\phi^*(g)= \varphi(z)|dz|^2$, $\varphi(z)>0$).
Prove the Riemann-Roch theorem for compact Riemann surfaces (Theorem 16.9 in [1]). Here the genus of the Riemann surface $X$ is defined as the dimension of $H^1(X, {\mathcal O})$, where ${\mathcal O}$ is the sheaf of holomorphic functions on $X$. Then prove that the genus equals half of 1st Betti number (Theorem 19.14 in [1].) This step one can, in principle, shorten by proving that if $X$ is a simply-connected Riemann surface, $H^1(X, {\mathcal O})=0$: Use the Dolbeault isomorphism and the fact that every holomorphic 1-form $\omega$ on a simply-connected surface $X$ is exact, $\omega= df$, $f=\int_{z_0}^z \omega$. (See [1], Theorem 15.14.)
Use the Riemann-Roch theorem to prove that if $X$ is a Riemann surface of has genus $0$ then $X$ is biholomorphic to $CP^1$ as it admits a degree 1 holomorphic map to it. (Corollary 16.13 in [1].)
Lastly, if $M$ is a smooth compact simply-connected surface, it hs to be oriented; picking an arbitrary Riemannian metric on $M$ converts it to a Riemann surface $X$ as above. By Hurewitz theorem, $b_1(M)=0$. Since $X$ is biholomorphic to $S^2$, $M$ is diffeomorphic to $S^2$.
Needless to say, this proof is much more complicated than the standard one (still, simpler than the one via the Ricci flow, although both are analytical in nature).
[1] O.Forster, "Lectures on Riemann surfaces." Springer Verlag, 1981.