Let $(f_n)\subseteq\operatorname{Lip}(X,d)$ be a Cauchy sequence. In particular, $(f_n)$ is a Cauchy sequence of continuous functions with respect to the sup norm, so it converges uniformly to a (continuous) function $f$ (this is what you already knew). Let's show that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.
Given $\varepsilon>0$, choose $N\in\mathbb{N}$ such that $\Vert f_n-f_m\Vert<\varepsilon$ whenever $n,m\geq N$. Given $n,m\geq N$ and $x\neq y$ in $X$, we have
$$\frac{|f_n(x)-f_m(x)-(f_n(y)-f_m(y))|}{d(x,y)}\leq \Vert f_n-f_m\Vert_d\leq \Vert f_n-f_m\Vert<\varepsilon.$$
Letting $m\rightarrow \infty$ and taking the sup on $x\neq y$, we obtain that $\Vert f_n-f\Vert_d<\varepsilon<\infty$ for any $n\geq N$, so (given one such $n$), we have $f_n-f\in\operatorname{Lip}(X,d)$, thus $f=f_n-(f_n-f)\in\operatorname{Lip}(X,d)$. Furthermore, we have just proved that $\Vert f_n-f\Vert_d\rightarrow 0$ as $n\rightarrow\infty$, and we already knew that $\Vert f_n-f\Vert_\infty\rightarrow 0$ as $n\rightarrow\infty$. This means exactly that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.
Therefore, $\operatorname{Lip}(X,d)$ is a Banach space.
Now, let $f,g:X\rightarrow\mathbb{R}$. Let $x\neq y$ in $X$. Then
$$|(fg)(x)-(fg)(y)|\leq|f(x)||g(x)-g(y)|+|f(x)-f(y)||g(y)|$$
$$\leq\Vert f\Vert_{\infty}\Vert g\Vert_d d(x,y)+\Vert f\Vert_d d(x,y)\Vert g\Vert_\infty.$$
So given any $z\in X$,
$$|(fg)(z)|+\frac{|(fg)(x)-(fg)(y)|}{d(x,y)}\leq\Vert f\Vert_\infty\Vert g\Vert_\infty+\Vert f\Vert_\infty\Vert g\Vert_d+\Vert f\Vert_d\Vert g\Vert_\infty$$
$$\leq (\Vert f\Vert_\infty+\Vert f\Vert_d)(\Vert g\Vert_\infty+\Vert g\Vert_d).$$
Taking the sup on $x\neq y$ and $z\in X$, we obtain $\Vert fg\Vert\leq\Vert f\Vert\cdot\Vert g\Vert$.
Remark: Maybe this exercise can be easier if you first show that $\Vert f\Vert_d=\inf\left\{K\geq 0:\forall x,y\in X,\ |f(x)-f(y)|\leq Kd(x,y)\right\}$, and then use that expression.
Best Answer
I will use $a$ instead of $\alpha$. Let $T$ be the interval $[0,2\pi]$. Let $f$ be a function of the shape
We identify $f:T\to \Bbb R$ (with $f(0)=f(2\pi)$) with a $2\pi$-periodic function on $\Bbb R$.
The quantity inside the supremum defining $r(f)$ is bounded, because (we may and do assume $h>0$, and then) the function $$ \begin{aligned} T^2&\to\Bbb R\ ,\\ (t,h)&\to\frac 1{h^a}(f(t+h)-f(t))&&\text{ for }t\in T\ ,\ h\in(0,2\pi]\ ,\\ (t,0)&\to\lim_{h\searrow0} \frac 1{h^a}(f(t+h)-f(t))=\delta_0(t)&&\text{ for }t\in T\ , \end{aligned} $$ is bounded. (It is continuous on a compact set obtained from $T^2$ obtained by removing a "small open (quarter)disc" with center in $(0,0)$. Then the control around $(0,0)$ comes from $(t+h)^a-t^a\le h^a$. This holds since for $t=0$ or $h=0$ this is clear, and else by $a$-homogenity it is enough to show $(1+h)^a\le 1+h^a$, i.e. $1+h\le 1+\frac 1ah^a\le (1+h^a)^{1/a}$.)
So $f$ is an element in the given Banach space. Let now $\tau>0$ be the number used for the shift of $f$. The shifted function $f_\tau$ is zero in a neighborhood of $0$, so $f-f_\tau=f$ in this small neighborhood. So for the special $t=0$ we have $$ \begin{aligned} r(f_\tau-f) &\ge \sup_{0< h}\frac{|(f_\tau-f)(0+h)-(f_\tau-f)(0)|}{h^a} \\ &= \limsup_{0< h}\frac{|f(0+h)-f(0)|}{h^a} \\ &= \sup_{0< h}\frac{h^a}{h^a} \\ &= 1\ . \end{aligned} $$