First recall the pointwise order: for sets $A,B$,
$$A \leq B :\equiv (\forall a \in A, b \in B :: a \leq b)$$
if either is a singleton set $\{x\}$ we write just $x$ for brevity;
eg, we define upper and lower bounds as
$$A_+ := \{ u \,:\, A \leq u \} \;\;,\;\; A_- := \{l \,:\, l \leq A\}$$
Anyhow, to the task at hand,
$
\;\;\; \text{sups exist} .
\\ \equiv \text{for every set } A \text{ there is an element } s
\text{ being the least upper bound of } A
\\\equiv \forall A :: \exists s :: s \text{ upper bound of A and least such}
\\\equiv \forall A :: \exists s :: \; A \leq s \leq A_+
\\\Rightarrow \forall A :: \exists s :: \; A_- \leq s \leq A_{-+} \;
\text{by instantiation: it holds for all $A$, and so holds for $A_-$}
\\\Rightarrow \forall A :: \exists s :: \; A_- \leq s \leq A \;
\text{ since $A \subseteq A_{-+}$ and $X \subseteq Y \geq l \Rightarrow l \leq X$}
\\ \equiv \text{for every set } A \text{ there is an element } s
\text{ being a lower bound of } A \text{ and the greatest such }
\\\equiv \text{infs exist.}
$
Hence, existence of sups implies existence of infs. The converse holds by duality, (or an exercise to the reader.)
This' essentially Freeze_S's answer in a more linear form; aiming for clarity.
Hope this helps!