What jdods wrote in the comments essentially covdrs it, but here I make things very explicit.
The supremum of a set is the smallest upper bound and the infimum is the largest lower bound. This is the formal definition that holds for every ordering.
Let's decode this. There is a usual ordering $\leq$ on $\mathbb{R}$, that one extends to the extended real line by letting $-\infty\leq r\leq$ for each real number $r$. If $A$ is a set of extended real numbers (in particular, a set of real numbers), then $u$ is an upper bound of $A$ if $r\leq u$ for all $r\in A$ and $l$ is a lower bound of $A$ if $l\leq r$ for all $r\in A$. Note that $-\infty$ is a lower bound of every set and $\infty$ is an upper bound of every set. Moreover, every extended real number is both an upper and lower bound of the empty set, since the condition "for all $r\in\emptyset$" holds vacuosly.
A smallest element in a set of extended real numbers is a lower bound of the set that is also an element of the set. If a set has a smallest element, it is unique. The set $(0,1]$ has no smallest element. Similarly, a largest element in a set is an upper bound of the set that is actually in the set.
Now let $A$ be a set of extended real numbers with infimum $\infty$. Then $\infty$ is a lower bound of $A$ and is the largest such lower bound. If $r\in A$, then we must have $\infty\leq r$, since $\infty$ is a lower bound of $A,$ and this is only possible if $r=\infty$. So the only element $A$ can have is $\infty$. There are then two possibilities, either $A$ is nonempty, then we must have $A=\{\infty\}$, in which case $\infty$ is indeed a lower bound of $A$, and, as the only lower bound, is also the largest lower bound. The other possibility is that $A=\emptyset$. Now every extended real number is a lower bound of the empty set, so the largest lower bound is the largest extended real number, namely $\infty$. So these are all possibilities for $\infty$ to be the infimum of a set of extended real numbers.