I'm studying a theorem whose proof involves an application of the Gauss-Bonnet theorem to a connected closed Riemannian $2$-manifold $(M,g)$, and the proof seems to suggest that the Euler characteristic $\chi(M)$ of $M$ is equal to $2$. Is this a fact that can be proved? If so, how do I prove this fact without any basic knowledge of algebraic topology? Thank you.
Definition. A closed manifold is a compact manifold without boundary.
Edit. I'm actually working on the Geroch monotonicity about the Hawking quasi-local mass. If we denote the mean curvatures of a family of connected closed Riemannian $2$-manifolds $\Sigma_t$ by $H$, then the proof to the Geroch monotonicity involves a step as follows: $$\begin{align} \frac{d}{dt}\int_{\Sigma_t}H^2 d\mu_t&=4\pi\chi(\Sigma_t)+\int_{\Sigma_t}\left(-2H^{-2}|\nabla H|^2-R-\frac{1}{2}(\lambda_1-\lambda_2)^2-\frac{1}{2}H^2\right)d\mu_t\\ &\leq 8\pi-\frac{1}{2}\int_{\Sigma_t}H^2 d\mu_t \end{align}$$ Did I misunderstand the inequality? Was it implying that $\chi(\Sigma_t)\leq 2$ instead? I know exactly where the term $-\frac{1}{2}\int_{\Sigma_t}H^2 d\mu_t$ came from, so I hypothesized that $\chi(\Sigma_t)=2$ at first.
Addressing your edit: the author does not use the fact that $\chi(M) = 2$, but rather uses the fact that $\chi(M) \leqslant 2$. Indeed, for a connected orientable (two sided) surface, the Euler characteristic is $2-2g$, where $g \geqslant 0$ is the genus. Another way to prove this inequality is to use Betti numbers: indeed, $\chi(M) = \beta_0 - \beta_1 + \beta_2$, with $\beta_0 = \beta_2 = 1$ by the connected and orientable assumptions.
By the way, you are talking about Riemannian surfaces, but the Euler characteristic is a topological invariant, independent of any metric.