Are you used to Zorn's lemma proofs? Because here is a quick one:
Let $A$ be a torsion-free abelian group. A sub-monoid of $A$ is a subset $P$ containing $0$ and closed under addition. We'll say that $P$ is pointed if it does not contain both $x$ and $-x$ for any nonzero $x \in A$. Clearly, the union of an ascending collection of pointed sub-monoids is itself a pointed sub-monoid. So, by Zorn's lemma, there is a maximal element pointed sub-monoid of $A$.
Claim 1: For any $x \in A$, and any positive integer $k$, if $kx \in P$ then $x \in P$.
Proof: Let $Q = \{ x \in A : kx \in P\ \text{for some } k \in \mathbb{Z}_{>0} \}$. Note that $Q$ is a sub-monoid: It clearly contains $0$ and, if $k_1 x_1 \in P$ and $k_2 x_2 \in P$, then $k_1 k_2 (x_1+x_2) \in P$. We also claim that $Q$ is pointed: If there is $x \neq 0$ such that $k x \in P$ and $\ell(-x) \in P$ then $k \ell x \in P$ and $- k \ell x \in P$. Since $A$ is torsion-free, $k \ell x \neq 0$, contradicting that $P$ is pointed.
So $Q$ is a pointed monoid containing $P$ and, by the maximality of $P$, we have $Q=P$. In other words, if $kx \in P$ for $k>0$, then $x \in P$, as desired. $\square$
Claim 2: For any $x$ in $A$, either $x$ or $-x \in P$.
Proof: Suppose that $-x \not \in P$.
Let $P' = \{ y+nx : y \in P, n \in \mathbb{Z}_{\geq 0} \}$. Clearly $P'$ is a sub-monoid of $A$. We claim that $P'$ is pointed. If not, then we have $y_1 + n_1 x = - (y_2 + n_2 x)$ for some $y_1$, $y_2 \in A$ and some $n_1$, $n_2 \geq 0$ with $y_1+n_1 x \neq 0$.
So $y_1 + y_2 =- (n_1+n_2) x$. If $n_1=n_2=0$, then $y_1 = - y_2$ and, since $P$ is pointed, $y_1=y_2=0$. In this case, $y_1+n_1 x = 0$, while we assumed otherwise.
Alternatively, suppose that $n_1+n_2>0$. We know $y_1 +y_2 \in P$ and by claim 1, the identity $y_1 + y_2 =- (n_1+n_2) x$ then forces $-x \in P$, a contradiction.
So we now know that $P'$ is pointed. So $P'=P$ and $x \in P$, as desired. $\square$
With $P$ as above, define $y \leq z$ if $z-y \in P$. This is reflexive (as $0 \in P$) and transitive (as $P$ is closed under addition.) It is anti-symmetric (because $P$ is pointed) and a total order by the claim.
So we have equipped $A$ with a linear order. As $P$ is closed under addition, the order is compatible with the group structure, meaning that $w \leq x$ and $y \leq z$ means $w+y \leq x+z$.