Lemma 1   If $a,b,c \in R$ with $a<b$ then $a+c<b+c$

Proof. The contrapositive will be proven.

Let $a,b,c \in R$ with $a+c \ge b+c$ Note that $-c \in R$ Thus,

Lemma 2   If $|R| \neq 1$ and $R$ has a characteristic, then it must be $0$

Proof. Suppose not. Let $n$ be a positive integer such that $\operatorname{char}~R=n$ Since $|R| \neq 1$ it must be the case that $n>1$

Let $r \in R$ with $r>0$ By the previous lemma, $\displaystyle 0<r \le \ldots \le \sum_{j=1}^{n-1} r \le \sum_{j=1}^n r=0$ a contradiction.

Lemma 3   If $a,b \in R$ with $a \le b$ and $c \in R$ with $c<0$ then $ac \ge bc$

Proof. Note that $-c \in R$ and $0=c+(-c)<0+(-c)=-c$ Since $a \le b$ $-(ac)=a(-c) \le b(-c)=-(bc)$ Thus,

Lemma 4   Suppose further that $R$ is a ring with multiplicative identity $1 \neq 0$ Then $0<1$

Proof. Suppose that $0 \not< 1$ Since $R$ is an ordered ring, it must be the case that $1<0$ By the previous lemma, $1 \cdot 1 \ge 0 \cdot 1$ Thus, $1 \ge 0$ a contradiction.