Are we properly using mathematical induction?

First we have this principle: take any (mathematically well-formed) property $P(n)$, depending on $n$ if you can prove that

$$P(0)$$

$$\forall n \in \mathbb{N}\, P(n) \implies P(n+1)$$

from this you may conclude $$\forall n \in \mathbb{N}\, P(n)$$

From this induction principle you can derive many other induction principles: finite induction, backwards induction etc but we only need two of the following:

You may start from any other natural number other than $0$:

$$P(k)$$

$$\forall n \in \mathbb{N_{\ge{k}}}\, P(n) \implies P(n+1)$$

$$\forall n \in \mathbb{N_{\ge{k}}}\, P(n)$$

and a principle of strong induction:

Suppose you prove

$$\forall n \in \mathbb{N_{\ge{k}}}\,\Bigl(\forall k \in \mathbb{N_{\ge{k}}}\, [k < n \implies P(k)]\Bigr) \implies P(n)$$

then you may conclude

$$\forall n \in \mathbb{N_{\ge{k}}} P(n)$$

In words, the first principle of mathematical induction works like this: You prove the base case (it is absolutely necessary to do this!) and after that you prove that the property holds for some number $n$ by knowing only that it holds for the previous number $n-1$! And if you work with the principle of strong induction not only that you don't need to prove the base case - you can also consider that the property holds for all of the previous numbers! How cool is that? And this is exactly what you did in your proof by strong induction:

Let's suppose that $P(n)$ means than $n$ can be rewritten as sum of $2$'s and $3$'s.

So you take any number $n \ge 8$. You assume that forall $8 \le k < n$ that property holds and then you prove that it also holds for $n$. You do this by cases:

if $n = 8$ then $P(n)$

if $n = 9$ then $P(n)$

if $n = 10$ then $P(n)$

if $n > 10$ then $8 \le n - 3 < n$ and thus by assumption $P(n-3)$ from this follows $P(n)$ since it only has one more $3$.

So you see that $P(n)$ holds in all cases thus by principle of strong induction you may conclude $\forall n \in \mathbb{N_{\ge{8}}}\, P(n)$

Now let's see how you could proceed by using first (ordinary) induction:

The idea is to use ordinary induction 3 times to prove that the property holds for 3 different sequences of natural numbers, that is for $8 + 3n$, $9+3n$, $10+3n$ $n \ge 8$ from this you will be able to conclude that it in facts hold for any natural number greater than $8$, since it'll definitely be in one of those sequences.

I'll show how to do that for the first one, the other two are proved verbatim.

We prove $$\forall n \in \mathbb{N} \mbox{ we have } P(8+3n)$$

$$n = 0, \mbox{ we certainly have } P(8)$$

Now assume we have $P(8+3k)$ for some $k \in \mathbb{N}$ then we need to prove that $P[8 + 3(k+1)]$.

But $8+3(k+1) = (8 + 3k) + 3$ and since $8 + 3k$ could be rewritten as a sum of $3$'s and $2$'s then $(8+3k)+3$ can be also rewritten this way since it is the same but has one more 3. Thus, we have $P[8 + 3(k+1)]$ and thus by the ordinary induction you may conclude $$\forall n \in \mathbb{N}\, P(8+3n)$$

Note that $P(N)$ does not imply $P(N+1)$ in any immediate way (e.g. it is not immediately clear why $17$ should have the property just because $16$ has the property ... though see the inductive proof I give a little later in this post).

So, if you have to use weak induction, the obvious thing to do is what your professor does, and do $3$ separate weak inductive proofs:

1. Show that $P(8)$, and show that for all $n$: $P(8+3n)\Rightarrow P(8+3(n+1))$

2. Show that $P(9)$, and show that for all $n$: $P(9+3n)\Rightarrow P(9+3(n+1))$

3. Show that $P(10)$, and show that for all $n$: $P(10+3n)\Rightarrow P(10+3(n+1))$

Combined together, these proofs show that for all $n \ge 8$: $P(n)$

Another possible weak inductive proof is this:

Show that $P(8)$, $P(9)$, and $P(10)$ hold. Now suppose $P(N)$ holds with $N \ge 10$. Because $N \ge 10$, we know that to form $N$, we need to either use at least one $5$, or (if we don't use any $5$'s at all), at least three $3$'s. But if $P(N)$ is formed using at least one $5$, then we can form $P(N+1)$ by taking away one $5$ and adding two $3$'s. And if we use at least three $3$ to form $P(N)$, then we can form $P(N+1)$ by taking away three $3$'s and adding two $5$'s. So, either way, we can form $P(N+1)$

Your method uses strong induction, and is of course a perfectly good mathematical proof as well ... but maybe your professor had insisted on you using weak induction?

If the property is true for all $k$ such that $$8\le k\le n,$$ where $10\le n$, then it is true for $n-2$ and is true for $n+1$, and for all $k$ such that

$$8\le k\le n+1.$$