if $\delta(q, a) = q$ for all symbols $a$, show that $\delta^* (q, w) = q$ is true for all strings $w$
By popular request, here is what the proof could look like:


  By induction over the length of $w$.  Assume $\delta(q,a)=q$ for all
  $a$. Take any $w$. There are two cases:
  
  
  If $|w|=0$ then $w=\epsilon$ (the empty word). Now the definition of $\delta^*$ has "$\delta^*(q,\epsilon)=q$" without any assumption
  on $\delta$ or $q$. We thus have $\delta^*(q,w)=q$ in this first case.
  If $|w|&gt;0$ then $w$ is some $w'a$ and $|w'|&lt;|w|$. Now $\delta^*(q,w)=\delta^*(q,w'a)=\delta(\delta^*(q,w'),a)$ by definition
  of $\delta^*$. Since $|w'|&lt;|w|$ we have $\delta^*(q,w')=q$ by
  induction hypothesis. Finally $\delta^*(q,w)=\delta(q,a)$, which is
  $q$ by the assumption. We have proved the required conclusion in this
  second case too.
  
  
  All cases have been covered. The proof is complete.
  
  Note that the two cases can be considered in any order we like. And
  that we do not use that $|w'|=|w|-1$, just that $|w'|&lt;|w|$.


As you can see, the above proof uses the same arguments as in your solution, and at the same places. It is slightly more rigorous on the quantifications ("take any $w$" is a "$\forall w$" and "$w$ is some $w'a$" is a "$\exists w'\exists a:w=w'a$") and their nesting ($w'$ and $a$ depend on $w$).

It is the kind of proof I'd love to get from a bright student beginning automata theory. But it is too pedantic and pedestrian for experts who, in their own papers, will usually just state the conclusion as obvious from the assumption. They may even not state the conclusion at all and use it anyway.