Let $A$ and $B$ be nonempty bounded subsets of $R$ and let $A+B$ be the set of all sums $a+b$ where $a\in A$ and $b\in B$.
Prove $\inf(A+B)=\inf(A)+\inf(B)$
My attempt: Since $A$ and $B$ are nonempty and bounded, $\inf(A)$ and $\inf(B)$ exist.
First, we want to show that $\inf(A)+\inf(B)$ is a lower bound of $A+B$:
$inf(A)$ is a lower bound of $A$, so $a\geq\inf(A)$ for every $a\in A$ &
$inf(B)$ is a lower bound of $B$, so $b\geq\inf(B)$ for every $b\in B$
Next we will want to demonstrate that $\inf(A)+\inf(B)$ is the GREATEST lower bound. (This second part is where I must be missing a concept.)
Assume to the contrary that there exists $t$ which is a greatest lower bound for $(A+B)$ but with $t>\inf(A)+\inf(B)$. Then $t-\inf(B)>\inf(A)$, so in particular $t-\inf(B)$ is not a lower bound for A. So there exists $a\in A$ with $t-\inf(B)<a$, or $t<a+\inf(B)$, or $t-a<\inf(B)$. Notice that $t$ is not a lower bound for $B$. Thus, there exists $b\in B$ with $t-a>b$, or $t>a+b$. However $t$ is a lower bound for $A+B$ producing a contraction. Thus, $\inf(A+B)=\inf(A)+\inf(B)$.
$\Box$
The first part of your proof is correct, but I think there is a flaw in the second part. From $ t−a<\inf(B) $ you cannot conclude that there exists $b \in B$ with $t−a>b$. (Also "$t$ is not a lower bound for $B$" need not be correct if $A$ contains negative numbers.)
One possible way to prove that $\inf(A)+\inf(B)$ is the greatest lower bound for $A+B$ is:
Let $t$ be a any real number such that $t>\inf(A)+\inf(B)$. (We want to show that $t$ is not a lower bound for $A+B$.) Then $\epsilon := t - \inf(A) - \inf(B)$ is greater than zero.
Then $a + b \in A + B$ and $a + b < \inf(A) + \inf(B) + \epsilon = t$, so that $t$ is not a lower bound for $A+B$.