A valuation ring is local

Let $\mathfrak{m}:=\{x\in V:x=0\text{ or }{x}^{-1}\mathrm{\notin }V\}$. By [A1], an element of $V$ lies outside $\mathfrak{m}$ exactly when it is a unit, so $1\mathrm{\notin }\mathfrak{m}$ and $\mathfrak{m}$ is proper.

If $r\in V$ and $x\in \mathfrak{m}$, then $rx\in \mathfrak{m}$: if $rx\mathrm{\ne }0$ and $(rx{)}^{-1}\in V$, then $x^{-1}=r(rx{)}^{-1}\in V$, contradicting $x\in \mathfrak{m}$.

Let $x,y\in \mathfrak{m}$. If $x+y\mathrm{\notin }\mathfrak{m}$, then $x+y$ is a unit by step 1.1. If $x=0$ or $y=0$ this contradicts $x,y\in \mathfrak{m}$, so assume $x,y\mathrm{\ne }0$. By [F1], either $y\mathrm{/}x\in V$ or $x\mathrm{/}y\in V$; in the first case $x^{-1}=(x+y{)}^{-1}(1+y\mathrm{/}x)\in V$, and in the second case $y^{-1}=(x+y{)}^{-1}(1+x\mathrm{/}y)\in V$, again a contradiction. Thus $x+y\in \mathfrak{m}$.

Steps 2.1 and 2.2 show that $\mathfrak{m}$ is an ideal. Every proper ideal contains no unit, so every proper ideal is contained in $\mathfrak{m}$. Hence $\mathfrak{m}$ is the unique maximal ideal of $V$, and $V$ is local.