The gauss map $ N$ is surjective if the surface is compact

For (2), the idea is to use the inverse function theorem. Take $v \in S^2$. Then, by part (1), there is some $w \in M_+$ such that $N(w) = v$. By the IFT, there are neighborhoods $U_w$ of $w$ and $U_v$ of $v$ such that $N\mid_{U_w} : U_w \to U_v$ is a diffeomorphism. Note $S^2 = \cup_v U_v$, so by compactness, $S^2 = \cup_{i=1}^n U_{v_i}$ for some $v_1,\dots,v_n \in S^2$. We can assume the sets $U_{v_1},\dots,U_{v_n}$ are disjoint (explanation below). Then, the sets $U_{w_1},\dots U_{w_n}$ are disjoint since $N \mid_{U_{w_i}}$ is a bijection for each $i$. Therefore, $4\pi = \int_{S^2} dA = \sum_{i=1}^n \int_{U_{v_i}} dA = \sum_{i=1}^n \int_{U_{w_i}} |\det dN| dA \le \int_{M_+} |K| dA$.

To see why we can assume $U_{v_1},\dots, U_{v_n}$ are disjoint, first note that replacing $U_{v_i}$ by two subsets whose union is $U_{v_i}$ is permissible, since the restriction of a diffeomorphism is a diffeomorphism. What we do is replace $U_{v_i}$ by $U_{v_i}$ intersected with all possible combinations of complements of $U_{v_1},\dots U_{v_{i-1}},U_{v_{i+1}},\dots,U_{v_n}$. It's hard to word this, so here's an example. Suppose we have three sets, $A,B,C$. We replace $A$ by $A \cap B \cap C, A \cap B^c \cap C, A \cap B \cap C^c, A \cap B^c \cap C^c$, and similarly with $B$ and $C$. It's easy to see that any pair of sets in this new collection (which has 12 elements) will be disjoint.