A germ is a unit exactly when its value at $0$ is nonzero, so ${\mathcal{O}}_{m,0}$ is local

Suppose $f(0)\mathrm{\ne }0$. Choose a representative, still called $f$, on a neighbourhood $U$ of $0$. By continuity from [L2], after shrinking $U$ we have $f(z)\mathrm{\ne }0$ for every $z\in U$. Then [L2] makes $1\mathrm{/}f$ holomorphic on $U$, so $[f][1\mathrm{/}f]=[1]$ in ${\mathcal{O}}_{m,0}$. Hence $[f]$ is a unit.

Suppose $f(0)=0$. For any germ $[g]\in {\mathcal{O}}_{m,0}$ one has $(fg)(0)=f(0)g(0)=0$, so $[f][g]\mathrm{\ne }[1]$. Therefore $[f]$ is not a unit.

Steps 1.1 and 1.2 show that the nonunits are exactly the germs vanishing at $0$, namely the elements of ${\mathfrak{m}}_{m,0}$ from [L1]. Any proper ideal contains no unit, so every proper ideal of ${\mathcal{O}}_{m,0}$ is contained in ${\mathfrak{m}}_{m,0}$. Since $1\mathrm{\notin }{\mathfrak{m}}_{m,0}$, this ideal is proper and therefore the unique maximal ideal. By [L3], ${\mathcal{O}}_{m,0}$ is local.