It’s not very well organized, and it has some extraneous clutter, but it also has the core of the argument. You want to show that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x,c\in\operatorname{dom}f$ and $|x-c|<\delta$, so in a polished version of the argument your first step should be:
Suppose that $f$ is Lipschitz continuous on some set $S$ with Lipschitz constant $M$, and fix $\epsilon>0$.
You’ve already worked out that $\epsilon/M$ will work for $\delta$, so you can even start out with:
Fix $\epsilon>0$ and let $\delta=\frac\epsilon{M}$.
Now you want to show that this choice of $\delta$ does the job.
Clearly $\delta>0$. Suppose that $x,c\in S$ and $|x-c|<\delta$. Then by the Lipschitz continuity of $f$ we have $|f(x)-f(c)|\le M|x-c|<M\delta=\epsilon$, so $f$ is uniformly continuous on $S$. $\dashv$
Added: Your heuristic interpretation of Lipschitz continuity is inaccurate enough that it may well lead you astray at some point. Consider the function
$$f(x)=\begin{cases}
x\sin\frac1x,&\text{if }x\ne0\\
0,&\text{if }x=0\;.
\end{cases}$$
This function has no vertical asymptotes, but it’s not Lipschitz continuous:
$$\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}=\frac{2n\pi}{\pi/2}=4n\;,$$
which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short.
You can take $K=1$ in the definition.
For all $x_1,x_2\in X$, we have: $$d(x_1,a)\leq d(x_1,x_2)+d(x_2,a)$$ and $$d(x_2,a)\leq d(x_1,x_2)+d(x_1,a)$$ by the triangle inequality. We may rewrite these as $$d(x_1,a)-d(x_2,a)\leq d(x_1,x_2)$$ and $$d(x_2,a)-d(x_1,a)\leq d(x_1,x_2).$$ These two inequalities together mean precisely that $$|d(x_1,a)-d(x_2,a)|\leq d(x_1,x_2)$$ which is Lipschitz continuity of $f$ with constant $K=1$.
For the first question, you are correct. $S$ is open because $S=f^{-1}(r,\infty)$ and $f$ is continuous.
Best Answer
$f$ is K-Lipschitz continuous implies $f_i$ is Lipschitz continuous $\forall i \in \{1, \dots, n\}$:
$\left | f_i(x)-f_i(y) \right |\leq \left \| f(x)-f(y) \right \|\leq K\left \| x-y \right \|$
Each coordinate $f_i$ is $K_i$-Lipschitz continuous implies $f$ is Lipschitz-continuous:
$f=\sum_{i=1}^{n}f_i\mathbf{e_i}$ where $\mathbf{e_i}=\left (0,...,0,1,0,...,0 \right )$ where the $1$ is at the $i$-th position.
Note that $\left \| f_1(x)\mathbf{e_1}+f_2(x)\mathbf{e_2}-f_1(y)\mathbf{e_1}-f_2(y)\mathbf{e_2} \right \|\leq \left \| f_1(x)-f_1(y) \right \|+\left \|f_2(x)-f_2(y) \right \|\leq L_1\left \| x-y \right \|+L_2\left \| x-y \right \|\leq (L_1+L_2)\left \| x-y \right \|$
The result follows by induction.