For this, taking an example doesn't prove anything. You can use counterexamples when you want to disprove a given statement.
To show this one by one
Let $s\in A$. Clearly, the first letter of any $s$ is equal to the first letter of $s$, and similarly for the last letter. Hence, $sRs$ and the relation is reflexive.
Let $s,t\in A$ such that $sRt$. By definition, the first or last letter of $t$ is the same as $s$. So, the first or last letter of $s$ must be the same as $t$, hence $tRs$, and the relation is symmetric.
Let $r,s,t\in A$ such that $rRs$ and $sRt$. Then, at least one out of the first and last letters of $r$ and $s$ are the same, and one out of the first and last letters of $s$ and $t$ are the same. But it is not necessary that this hold for $r$ and $t$. Let $r_1$ be the first letter of $r$ and $r_4$ be the lat letter. Suppose $r_1=s_1$ and $r_1\ne t_1$, and $s_4=t_4$ and $r_4\ne t_4$. Clearly, $rRt$ is not true. For example, let $r=aaab$, $s=aaaa$, $t=baaa$. Hence the relation is not transitive.