The idea is to "back off the inf's" and use a "since $\epsilon$ is arbitrary" argument. More precisely,
$1).\ A$ is bounded below, which implies that $A+k$ is also. This means that each has a finite greatest lower bound.
$2).\ $ Set $\alpha=\inf A$ and $\beta=\inf (A+k)$. You want to show that $\beta=\alpha+k.$ As you rightly point out, if we can show $\beta\le\alpha+k$ and $\beta\ge\alpha+k,$ we will have proved the claim.
$3).\ $ There is an $a\in A$ such that $a<\alpha+\epsilon.$ Then, $a+k\in A+k$ and so $\beta\le a+k<\alpha+k+\epsilon.$ Since $\epsilon>0$ is arbitrary, we get $\beta\le\alpha+k$.
$4).$ There is a $b\in A+k$ such that $b\le \beta+\epsilon.$ But $b=a+k$ for some $a\in A$ (by definition of the set $A+k$) and of course $\alpha\le a.$ Therefore, $\alpha+k\le a+k=b< \beta+\epsilon$ and so by the same reasoning as in $(3)$, we get $\alpha +k\le \beta.$