Let $A$ be a nonempty set. Show that if $R$ is a symmetric and transitive relation on $A$ such that $rngR = A$, then $R$ is reflexive on $A$.
So I proved this by saying: For all $x,y\in A$, $(x,y)\in R$ implies $(y,x)\in R$. (definition of symmetric relation)
And for all $x,y\in A$, $(x,y)\in R$ and $(y,x)\in R$ implies $(x,x)\in R$. (definition of transitive relation)
Therefore, $R$ is reflexive.
My prof made a comment on my proof, asking what rngR=A is for, which I don't know so an answer would be very much appreciated.
Suppose that $\operatorname{rng}R\ne A$; then there is some $x\in A$ that is not in the range of $R$. But then $R$ cannot be reflexive, because $\langle x,x\rangle\notin R$.
The problem with your argument is that in order to show that $\langle x,x\rangle\in R$, you have to have some $y\in A$ such that $\langle x,y\rangle\in R$ and hence by symmetry $\langle y,x\rangle\in R$. If you don’t have such a $y$ in the first place, you can’t use transitivity to get $\langle x,x\rangle$. And if $x$ is not in the range of $R$, then you don’t have such a $y$.
If $\operatorname{rng}R=A$, then for each $x\in A$ there is some $y\in A$ such that $\langle y,x\rangle\in R$. Symmetry then tells you that $\langle x,y\rangle\in R$, and transitivity that $\langle x,x\rangle\in R$, so in fact we see that $R$ is reflexive if and only if $\operatorname{rng}R=A$.