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 $A$.
