Proving that $s$ is a reflexive closure of $r$ I have this problem that I cannot figure out. Could you please help me out with this?
Let $r$ be a relation on the set $X$ and let $$R:=\{t:t \text{ is a  reflexive relation on } X \text{ with } r\subseteq t \}.$$ For $x,y \in X$ we have $x \, s \, y$ iff for each $t\in R$ we have $x \, t \, y$. Show that $s$ is the reflexive closure of $r$.
My idea so far was as follows.
Suppose $x,y \in X$ and $x \ s \ y$. It follows that for all $t\in R$ we have $x \ t \ y$. Since $r\subseteq t$ for all $t\in R$, it follows $r \subseteq s$ and also $s \subseteq t$. Since each $t\in R$ is reflexive, $s$ is also reflexive.
Could you please verify my idea? I'm sure it has some mistakes.

Yeah, this has some problems.

Since $r\subseteq t$ for all $t\in R$, it follows $r \subseteq s$ and also $s \subseteq t$.

How does it follow that $r \subseteq s$? That's not clear.
Fortunately, to show that $s$ is reflexive we don't need it ... yes, we need that $s$ is reflexive, but given its definition that's the easy part. 
The harder part is that at some point you  need to show that $s$ is the closure of $r$, i.e. that $s$ is the smallest reflexive relation for which $r \subseteq s$. That part you haven't addressed at all yet.
