Showing that Dedekinds Theorem $\implies$ Axiom of Completeness 
Axiom Of Completeness: If $X$ and $Y$ are nonempty subsets of $\mathbb{R}$ having the property that $x\leq y$ for every $x\in X$ and $y\in Y$ then there exists a $c\in\mathbb{R}$ such that $x\leq c\leq y$ for all $x\in X$ and $y\in Y$.
Dedekinds Theorem (or part d) ): If $X$ and $Y$ are sets defined in $c)$, then either $X$ has a maximal element, or $Y$ has a minimal element.
Apologies for the confusion; I did not know there were multiply versions of the theorems above.
I am currently on part e); I have managed to show that the Axiom of Completeness implies Dedekind's theorem from pard d), but I am having trouble showing the Axiom of Completeness from Dedekind's Theorem.
My goal: To show that Dedekinds Theorem $\implies$ Axiom of Completeness
I assumed all the necessary properties that $X, Y$ are non-empty subsets of $\mathbb{R}$ and that $x\leq y$ for all $x\in X$ and $y\in Y$. I noticed in part c) that if $X\cup Y=\mathbb{R}$ then $\mathrm{sup}\ X=\mathrm{inf}\ Y$ which would be a really nice choice for $c\in\mathbb{R}$ to close off that $x\leq c\leq y$ (showing the axiom of completeness) however, I am not sure how to show that $X\cup Y=\mathbb{R}$. It's obvious that $X\cup Y\subset \mathbb{R}$(since they are both subsets), but the other inclusion is tripping me up; perhaps I went down a wrong path, or there is a trick to this. Any hints or directions?
 A: Ok thanks for the clarification. So let $X$ and $Y$ be nonempty sets satisfying the property that for all $x\in X$ and $y\in Y$, we have that $x\leq y$. We need to show there is some intermediate $c$ assuming that point $\textit{d}$ holds.
From point $\textit{b}$ we have that $\sup{X}\leq\inf{Y}$. Now let us define two auxiliary sets $\hat{X}$ and $\hat{Y}$ by
\begin{equation*}
\begin{split}
\hat{X} & :=\{z\in\mathbb{R}:z\leq\sup{X}\}\\
\hat{Y} & :=\{z\in\mathbb{R}:z\geq\inf{Y}\}
\end{split}
\end{equation*}
Clearly these are nonempty as $X\subset\hat{X}$ and $Y\subset\hat{Y}$ and we also have that for all $x\in\hat{X}$ and all $y\in\hat{Y}$ it holds that $x\leq y$.
$\textit{Case 1:}$ $\mathbb{R}\setminus(\hat{X}\cup\hat{Y})\neq\varnothing$. Then let $c$ be a real number in the difference. By definition then $x\leq \sup{X}\leq c\leq \inf{Y}\leq y$ for all $x\in X$ and $y\in Y$.
$\textit{Case 2:}$ $\mathbb{R}=\hat{X}\cup\hat{Y}$. Then by point $\textit{d}$, $\hat{X}$ has a maximal element, call it $c$ (analogous treatment for the case when $c$ is the minimum of $\hat{Y}$). Then $c=\sup{\hat{X}}\geq\sup{X}\geq x$ for all $x\in X$. But then by point $\textit{b}$ applied to the sets $\hat{X}$ and $\hat{Y}$, we know that $c=\sup{\hat{X}}\leq \inf{\hat{Y}}\leq\inf{Y}\leq y$ for all $y\in Y$.
For the sake of completeness, a certain version of Dedekind's theorem can be used to prove the existence of suprema and infima assuming that you don't want to assert their existence despite how you've presented the problem.
$\textit{Axiom:}$ Whenever $A$ and $B$ are non-empty sets of reals satisfying $a\leq b$ for all $a\in A$ and $b\in B$ and $\mathbb{R}=A\cup B$, then either $A$ has a maximum or $B$ has a minimum.
We use the above axiom to prove that a non-empty set $A$ that has an upper bound has a supremum.
Indeed, let $U(A):=\{y\in\mathbb{R}:a\leq y\text{ for all }a\in A\}$ and let $V(A):=\mathbb{R}\setminus U(A)$. Clearly $\mathbb{R}=U(A)\cup V(A)$. Now pick arbitrary $y\in U(A)$ and $x\in V(A)$. Since $x\in V(A)$, there must exist some $a_x\in A$ such that $x<a_x$. But $a_x\leq z$ for any $z\in U(A)$ by construction, therefore $x<a_x\leq y$. So we can apply the axiom to conclude that either $V(A)$ has a maximum or that $U(A)$ has a minimum and call this number $t$.
If $t=\min{U(A)}$ then $t$ is by definition the least upper bound of $A$ and therefore the sup of $A$. So assume wlog that $t=\max{V(A)}$. But then there exists a unique $a_t\in A$ such that $a_t>t$. The existence is given by the fact that $t$ is not an upper bound for $A$ and the uniqueness by the fact that if there were a different $a_{t'}\in A$ satisfying $a_{t'}>t$, then $\min \{a_t,a_{t'}\}>t$ would be a bigger element in $V(A)$. But then $a_t$ is clearly the maximum of the set $A$ and therefore the sup of $A$.
