Language is recursive, hence recursively enumerable
Depending on what definitions you use, the proof can be trivial.  We can all agree that a set&nbsp;$S$ is recursive if there is a Turing machine that halts on every input, accepts the members of&nbsp;$S$ and rejects the non-members of&nbsp;$S$. Such a Turing machine is called a decider of the set.

One definition of the recursively enumerable sets is that $S$&nbsp;is RE if there is a Turing machine that accepts every member of&nbsp;$S$ and does not accept any non-member.  That is, if $x\notin S$, the machine either rejects it or loops forever.  By this definition, it is trivial that every recursive language is RE: the machine that decides the language has the property that it accepts everything in the set and nothing outside it.

The other definition is that a set&nbsp;$S$ is RE if there is a Turing machine that accepts every member of the set and loops forever for every non-member.  This appears to be the definition your book is using.  Now, suppose you have a machine that decides a set&nbsp;$S$.  Modify it so that, instead of rejecting non-members of the set, it loops forever.  This gives a machine that accepts every member of&nbsp;$S$ and loops forever for every non-member.  This is exactly what you need to conclude that $S$&nbsp;is RE.

The same idea proves that the two definitions of RE are the same.  Any machine that obeys the second definition also obeys the first one (the first says it can either loop or reject for non-members: it happens to loop for all of them but that's a valid choice).  Any machine that obeys the first definition can be converted to obey the second by the same trick: replace rejection with looping.