I'm attempting to prove the following and I think I have managed to prove it, but since I'm quite new to real analysis, I would like a second set of eyes to verify my proof! Note that I have seen other proofs here proving that Lipschitz continuity of $\alpha=1$ gives uniform continuty, but I have not found this question being asked here, so I doubt it is a repost
Let $\alpha$ be a fixed positive real number and let $I=(a,b)$ be an open interval where possibly $a=-\infty$ and $b=\infty$. A real-valued function $f(x)$ on $I$ is said to be Lipschitz continuous of exponent $\alpha$ if there is a constant $C$ such that for all $x, y \in I$ with $|x − y| \leq 1$ we have $|f(x) − f(y)| \leq C|x − y|^{\alpha}$
i) Prove that if $f(x)$ is Lipschitz continuous of exponent $\alpha>0$ then $f(x)$ is uniformly continuous on $I$.
ii) Prove that if $f(x)$ is differentiable and the derivative $f'(x)$ is bounded on $I$, then $f(x)$ is Lipschitz continuous of exponent $\alpha$=1.
$\textbf{Proof of i)}$ We are given that $\forall x,y \in I ; |x-y|\leq 1 \Rightarrow |f(x)-f(y)|\leq M|x-y|^{\alpha}$ Now, let $\delta=\text{min}\left\{1,\sqrt[\alpha]{\frac{\varepsilon}{M}}\right\}$ to ensure $|x-y|\leq 1 \Rightarrow |x-y|\leq \delta$. Suppose $\delta=1$, then, since we assume Lipschitz continuity $|x-y|\leq \delta \Rightarrow |f(x)-f(y)|\leq M \delta^\alpha\leq M$, then we have uniform continuity (we could rename $M=\varepsilon$ to be very explicit). Suppose now that $\delta=\sqrt[\alpha]{\frac{\varepsilon}{M}}$, then $|x-y|\leq \delta \Rightarrow |f(x)-f(y)|\leq M\delta^{\alpha}=M\frac{\varepsilon}{M}=\varepsilon$ and we are done.
$\textbf{Proof of ii)}$ We know the derivative $f'(x)$ exists at all $x\in I$ and that $-\infty < f'(x) < \infty$. Let $y\in I$ and let $f'(y)=C$, then since the limit $f'(y)=\lim_{x\rightarrow y} \frac{f(x)-f(y)}{x-y}$ exists, we know by definition that $\forall \varepsilon >0, \exists \delta >0 ; |x-y|\leq \delta \Rightarrow \left|\frac{f(x)-f(y)}{x-y}-C\right|\leq \varepsilon$, or $\forall \varepsilon >0, \exists \delta >0 ; |x-y|\leq \delta \Rightarrow \left|f(x)-f(y)\right|\leq (\varepsilon+|C|)|x-y|$. Choose $\varepsilon=\varepsilon_{0}$ such that $\delta=1$ and let $M=|C|+\varepsilon_0$ from which we have shown that $\forall x,y \in I, |x-y|\leq 1\Rightarrow \left|f(x)-f(y)\right|\leq M|x-y|^1$ and we are done.
Are my proofs correct?
First, a comment regarding your notation. $f(x)$ is not a function, but the function $f$ evaluated at the point $x$, hence, just a value. So not $f(x)$ is called to be Lipschitz continuous function but $f$ is. The same for differentiability, $f$ is supposed to be differentiable, not $f(x)$ (which is of course true, but not what we want).
Now, regarding your proofs: For (i), the case where you consider $\delta = 1$ is not quite correct. You cannot just rename $M=\varepsilon$ as $M$ is supposed to be fixed and $\varepsilon > 0$ is arbitrary. You don't even have to do a case distinction. Just say if $\vert x-y \vert < \delta$, then $\vert f(x) - f(y) \vert \leq M \vert x-y \vert^\alpha \leq M\delta^\alpha \leq \varepsilon.$
For (ii), it's not quite clear how you get from $\vert\frac{f(x)-f(x)}{x-y} - C\vert < \varepsilon$ to $\vert f(x) - f(y) \vert \leq (\varepsilon + \vert C\vert)\vert x-y \vert$. (I don't even think it's correct). I would suggest you take a look at the mean value theorem, from which (ii) follows immediately using that $f'$ is bounded.
Note that nowhere we made use of the assumption that $\vert x-y \vert \leq 1$. Your definition of Lipschitz continuity with exponent $\alpha$ is usually just called $\alpha$-Hölder continuity and defined without this additional assumption.