Let $a/b$ be a fraction in lowest terms with $0<a/b<1$. Prove that there exists $n∈\mathbb N$ such that $$\frac{1}{n+1}\le\frac{a}{b}<\frac{1}{n}$$
My proof:
If $n∈\mathbb N$ then $n<n+1$ and thus $$\frac{1}{n+1}<\frac{1}{n}$$ then if we take $n=b$ it follows that $$\frac{1}{n+1}=\frac{1}{b+1}<\frac{1}{b}\le\frac{a}{b}$$ so we have shown that $$\frac{1}{n+1}\le\frac{a}{b}$$
Your proof for part a is rather flawed: After doing a few correct manipulations, you go from the valid statement that $\frac{1}{n+1} < \frac{1}{n}$ to simply stating that which you were to have proven.
The sort of thing you would need to do is first show that there exists some $n$ such that $\frac{1}{n+1} < a/b$ (hint: choose $n=b$), then assume there is no $n$ satisfying $\frac{1}{n+1} \leq a/b < \frac{1}{n}$ and show on that basis by induction that $a/b \geq 1$. Having demonstrated that contradiction, you will have proven the theorem.