Answer
Proof. Let S be a nonempty set, and let $R=\emptyset$. First, we show R is symmetric on S. To show R=$\emptyset$ is symmetric on S, we must show $(a,b)\in R \implies (b,a) \in R$ for all $a,b \in S$. But since $R$ is empty, $(a,b) \in R$ is false for all $a,b \in S$. This means $(a,b) \in \emptyset \implies (b,a) \in \emptyset$ is a conditional sentence with a false antecedent for all $a,b \in S$. Thus this conditional is true for all $a,b \in S$. So the relation $R=\emptyset$ is symmetric on S.
Next, we show $R=\emptyset$ is transitive on S. To show this, we must show the conditional sentence $((a,b)\in \emptyset \wedge (b,c)\in \emptyset) \implies (a,c) \in \emptyset$ is true for all $a,b,c \in S$. But $(a,b)\in \emptyset$ and $(b,c) \in \emptyset$ are both false for all $a,b,c \in S$, which means the conjunction $(a,b)\in \emptyset \wedge (b,c) \in \emptyset$ is false for all $a,b,c \in S$. Thus the conditional sentence $((a,b)\in \emptyset \wedge (b,c)\in \emptyset) \implies (a,c) \in \emptyset$ has a false antecedent for all $a,b,c \in S$, and is, therefore, true for all $a,b,c \in S$. So $R=\emptyset$ is transitive on S.
Finally, we show the relation $R= \emptyset$ is not reflexive on S.
Since S is nonempty, there exists an element $a \in S$. But since $R=\emptyset$, $(a,a) \notin R$. Therefore, $R$ is not reflexive on S.
Work Step by Step
Step 1) Show R=$\emptyset$ is symmetric on the nonempty set S.
To show R=$\emptyset$ is symmetric, we must show $(a,b)\in R \implies (b,a) \in R$ for all $a,b \in S$. That is, we must show the conditional sentence $$(a,b)\in \emptyset \implies (b,a) \in \emptyset$$ is true for all $a,b \in S$. But notice $(a,b) \in \emptyset$ is false for all $a,b \in S$. So $$(a,b) \in \emptyset \implies (b,a) \in \emptyset$$ is a conditional sentence with a false antecedent for all $a,b \in S$, which means this conditional is true for all $a,b \in S$. So the relation $R=\emptyset$ is symmetric on S.
Step 2) Show $\emptyset$ is transitive on S
As in step one, we use the fact that a conditional sentence with a false antecedent is true.
To show $\emptyset$ is transitive, we must show the conditional sentence $$((a,b)\in \emptyset \wedge (b,c)\in \emptyset) \implies (a,c) \in \emptyset$$ is true for all $a,b,c \in S$. But $(a,b)\in \emptyset$ and $(b,c) \in \emptyset$ are both false for all $a,b,c \in S$, which means the conjunction $$(a,b)\in \emptyset \wedge (b,c) \in \emptyset$$ is false for all $a,b,c \in S$. This means the conditional sentence $$((a,b)\in \emptyset \wedge (b,c)\in \emptyset) \implies (a,c) \in \emptyset$$ has a false antecedent for all $a,b,c \in S$, and is, therefore, true for all $a,b,c \in S$. So $\emptyset$ is transitive on S.
Step 3) Show the relation $R= \emptyset$ is not reflexive on S.
To show $R$ is not reflexive, we must show there is an element in S that is not related to itself under $R$.
Now, since we know S is nonempty, we know there is at least one element in S, let's call it $a$. But since $R$ is empty, $(a,a) \notin R$. So $R=\emptyset$ is not reflex on S.
