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.
This is a classic application of Banach's fixed point theorem.
Banach fixed point theorem If $F:X\to X$ is such that there exists $\lambda\in [0,1)$ such that for every $x_1, x_2\in X$, $\|F(x_1) - F(x_2)\| \leq \lambda \|x_1 - x_2\|$, then there exists a unique fixed point of $F$.
Consider the function
$$ f(x) = \frac{1}{\mu} (y - Tx) $$
A solution to your original equation is also a fixed point of $f$. So if we can show that $f$ is a contraction mapping (equivalently it is Lipschitz with Lipschitz constant $<1$), then by Banach's fixed point theorem we are done.
But
$$ f(x_1) - f(x_2) = \frac{1}{\mu} [T(x_2) - T(x_1)] $$
so as long as you choose $\mu$ bigger than the Lipschitz constant of $T$ you have that $f$ is a contraction mapping.
Best Answer
Fix $\varepsilon>0$. We can find $N(\varepsilon)$ such that for all $n,m\geq N(\varepsilon)$ and all $x,y\in X$ with $x\neq y$: $\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{d(x,y)}\leq \varepsilon$. Since $f_n$ converges to $f$ pointwise, we have, taking the limit $m\to\infty$ for $x,y$ fixed that \begin{equation} (*) \quad \forall n\geq N(\varepsilon),\forall x,y\in X,x\neq y\quad\frac{|(f_n-f)(x)-(f_n-f)(y)|}{d(x,y)}\leq \varepsilon. \end{equation} In particular for $n=N(\varepsilon)$ we have $\frac{|f(x)-f(y)|}{d(x,y)}\leq \varepsilon+\frac{|f_{N(\varepsilon)}(x)-f_{N(\varepsilon)}(y)|}{d(x,y)}$, which shows that $f$ is Lipschitz. Now we have to show that $|f-f_n|_L$ converges to $0$. But this is a consequence of $(*)$, since we only have to take the supremum over these $x$ and $y$.
Note that we don't need to use Ascoli's theorem to show that $\{f_n\}$ has a limit (just use the same proof when you show that the set of the continuous real functions on a compact space for the uniform norm is a Banach space).
Here compactness is needed to be sure that each Lipschitz function is bounded, since such a function is continuous.