There are two Theorems in Chartrand's Graphs and Digraphs that follow the same line of argument and arrive at an inequality that I can't seem to understand where it is deduced from. The inequality being the following: $(deg(u)-1) + (deg(v)-1) \leq n-2$ and $deg(u) + deg(v) \leq n-2$ in Theorem $1.9$ and Theorem $2.5$ respectively.
The two Theorems are shown below:
Both proofs follow the same line of logic. I understand the proofs, however, how can we be certain that the sum of the degrees of vertices is always less than $(n-2)$?
For example, the deletion of $1$ from $deg(u)$ and $deg(v)$, respectively, from the LHS of the inequality: $(deg(u)-1) + (deg(v)-1) \leq n-2$ comes from there being no vertex $w$ adjacent to both vertex $u$ and $v$ (since the new graph does not contain a triangle), is that correct? Furthermore, how is it bounded above by $(n-2)$?
Also, in Theorem $2.5$, is it (implicitly) following a proof by contracdition method? Since it isn't stated, however comes to a contradiction.
Let $N_x$ denote the neighbourhood of vertex $x$ so that: $$ N_x = \{y \in V(G) \mid xy \in E(G)\} $$
For the inequality from Theorem 1.9, observe that $N_u \cap N_v = \varnothing$ (otherwise, if there was some $w \in N_u \cap N_v$, then $u \to w \to v \to u$ would be a triangle). Hence, observe that: \begin{align*} n &=|V(G)| \\ &\geq |\{u\} \cup \{v\} \cup N_u \cup N_v| \\ &= |N_u \cup N_v| &\text{since $u \in N_v$ and $v \in N_u$} \\ &= |N_u| + |N_v| &\text{since $N_u \cap N_v = \varnothing$} \\ &= \deg u + \deg v\\ \end{align*}
Likewise, for the inequality from Theorem 2.5, observe that $N_u \cap N_v = \varnothing$ (otherwise, if there was some $w \in N_u \cap N_v$, then $u \to w \to v$ would be a $(u, v)$-path). Hence, observe that: \begin{align*} n &=|V(G)| \\ &\geq |\{u\} \cup \{v\} \cup N_u \cup N_v| \\ &= |\{u\}| + |\{v\}| + |N_u| + |N_v| &\text{since all four sets are pairwise disjoint} \\ &= 2 + \deg u + \deg v\\ \end{align*}