First let us recall Zorn's lemma.
Zorn's lemma. Suppose that $(P,\leq)$ is a non-empty partial ordered set such that whenever $C\subseteq P$ is a chain, then there is $p\in P$ that for every $c\in C$, $c\leq p$. Then $(P,\leq)$ has a maximal element.
To use Zorn's lemma, if so, one has to find a partial order with the above property (every chain has an upper bound) and utilize the maximality to prove what is needed.
We shall use the partial order whose members are $(A,B)$ where $A,B$ are disjoint subsets of $\mathbb R^+$ each is closed under addition. We will say that $(A,B)\leq (A',B')$ if $A\subseteq A'$ and $B\subseteq B'$.
This is obviously a partial order. It is non-empty because we can take $A=\mathbb N\setminus\{0\}$ and $B=\{n\cdot\pi\mid n\in\mathbb N\setminus\{0\}\}$, both are clearly closed under addition and disjoint.
Suppose that $C=\{(A_i,B_i)\mid i\in I\}$ is a chain, let $A=\bigcup_{i\in I}A_i$ and $B=\bigcup_{i\in I} B_i$. To see that these sets are disjoint suppose $x\in A\cap B$ then for some $A_i$ and $B_j$ we have $x\in A_i\cap B_j$. Without loss of generality $i<j$ then $x\in A_j\cap B_j$ contradiction the assumption that $(A_j,B_j)\in P$ and therefore these are disjoint sets. The proof that $A$ and $B$ are closed under addition is similar.
Then $(A,B)\in P$ and therefore is an upper bound of $C$. So every chain has an upper bound and Zorn's lemma says that there is some $(X,Y)$ which is a maximal element.
Now all that is left is to show that $X\cup Y=\mathbb R^+$. Suppose that it wasn't then there was some $r\in\mathbb R^+$ which was neither in $X$ nor in $Y$, then we can take $X'$ to be the closure of $X\cup\{r\}$ under addition. If $X'\cap Y=\varnothing$ then $(X',Y)\in P$ and it is strictly above $(X,Y)$ which is a contradiction to the maximality. Therefore $X'\cap Y$ is non-empty, but then taking $Y'$ to be the closure of $Y\cup\{r\}$ under addition has to be disjoint from $X$, and the maximality argument holds again.
In either case we have that $X\cup Y=\mathbb R^+$.
Indeed we have to show that the set is closed under addition in each case. For example, consider $2/3$ and $1/3$. Both of them satisfy $|.|<1$ but $|2/3+1/3|$ is not strictly smaller than $1$. So c doesn't make a groupid either. The similar story can be seen for d for example. In fact $|3/2+(-4/2)|<1$ while $|3/2|\ge1,~~|-4/2| \ge1$.
Best Answer
It would help readability of your argument if you said something like this:
$1.$ Let $b$ be in $X$. Then $b-b$ is in $X$. So $0$ is in $X$.
$2.$ Because $0$ is in $X$, for any $b$ in $X$ we have $0-b$ is in $X$. So $-b$ is in $X$.
$3.$ For any $a$ and $b$ in $X$, $a+b=a-(-b)$, so by $(2)$, $a+b$ is in $X$.