Question: I think I proved this but I am not to confident in my proof. I am also not that good at formating my proofs so any feedback would be appreciated :)

Let $P(n)=9|10^n-1$

Base Case:
$9|10^1-1\implies9|9\checkmark$
So we have proven the base case.

Induction Step:
$$9|10^n-1\implies9k=10^n-1\implies10^n=9k+1$$
$$9|10^{n+1}-1\implies9|(9k+1)\cdot10-1\implies9|90k+9$$
Every term in the above has a common factor of 9 hence the left divides the right. We have thus shown that $P(n)\implies P(n+1)$ with a base case. Therefore by the principle of induction $P(n)$ holds $\forall n\in\mathbb{N}$.

**Edit:**
The second line of the induction step was flawed as it didn't really prove anything. The fixed version goes as follows:
$$10^{n+1}=(9k+1)\cdot10\implies10^{n+1}-1=90k+10-1\implies10^{n+1}-1=90k+9$$
We can see that this $10^{n+1}-1=90k+9$ is divisible by 9. We have thus shown that $P(n)\implies P(n+1)$ with a base case. Therefore, by the principle of induction, $P(n)$ holds $\forall n\in\mathbb{N}$.

Answer: 
I think you're almost all the way there-- the main feedback I'd give is around your final statement:

$$9|10^{n+1}-1\implies9|(9k+1)\cdot10-1\implies9|90k+9$$

Above you're assuming what you want to prove. Try working backwards starting with what you *know* to get to what you want to prove. Here, try the following to get started:

$$10^n = 9k + 1 \implies 10^{n+1} = (9k + 1) \cdot 10 \implies 10^{n+1} - 1 = ... $$

N.B. Also be sure to declare what $k$ is when you use it.

I hope this helps!
When you write $9\mid10^n-1\implies9k=10^n-1$, you should say what that $k$ is. Something like “$9k=10^n-1$” for some non-negative integer $k$.

And then your aim is to *prove* that $9\mid10^{n+1}-1$. That's not what you do. You *start* from $9\mid10^{n+1}-1$ and then you see what you can conclude from that.