Could someone please let me know if my solution is okay?
Let $A$ and $B$ be nonempty subsets of $\Bbb{R}$. Let $C = \left \{x + y : x \in A, y \in B\right \}$. Then if $A$ and $B$ have an infimum, then $C$ has the infimum $\operatorname{inf}(C) = \operatorname{inf}(A) + \operatorname{inf}(B)$.
$\operatorname{inf}(A) + \operatorname{inf}(B)$ is a Lower Bound For $C$
Let $z \in C$. Then $z = x + y$ with $x \in A$ and $y \in B$. Then by the definition of $\operatorname{inf}(A)$ and $\operatorname{inf}(B)$, $\operatorname{inf}(A) + \operatorname{inf}(B) \leq x + y$. Then $\operatorname{inf}(A) + \operatorname{inf}(B)$ is a lower bound for $C$.
$\operatorname{inf}(A) + \operatorname{inf}(B)$ is the Greatest Lower Bound For $C$
Since $C$ has a lower bound, $C$ has a greatest lower bound $\operatorname{inf}(C)$ such that $\operatorname{inf}(A) + \operatorname{inf}(B) \leq \operatorname{inf}(C)$. Let $\varepsilon > 0$. Then by the approximation property for infimum, there exists an $x$ in $A$ such that $x < \operatorname{inf}(A) + \varepsilon$ and there exists a $y$ in $B$ such that $y < \operatorname{inf}(B) + \varepsilon$. Adding the inequalities gives the following:
$\operatorname{inf}(C) \leq x + y < \operatorname{inf}(A) + \varepsilon + \operatorname{inf}(B) + \varepsilon = \operatorname{inf}(A) + \operatorname{inf}(B) + 2\varepsilon$
Then $\operatorname{inf}(C) < \operatorname{inf}(A) + \operatorname{inf}(B) + 2\varepsilon$ for every $\varepsilon > 0$. Then $\operatorname{inf}(C) < \operatorname{inf}(A) + \operatorname{inf}(B)$. Then $\operatorname{inf}(C) = \operatorname{inf}(A) + \operatorname{inf}(B)$.
Here is another way:
With $a \in A, b \in B$ we have $a+b \ge \inf A + \inf B$ hence $\inf C \ge \inf A + \inf B$.
With $a \in A, b \in B$ we have $a+b \ge \inf C$. If we take the $\inf$ over $a \in A$ we have $\inf a + b \ge \inf C$, and repeating for $b \in B$ we get the desired result.