If $f$ is locally Lipschitz on $X$ and $X$ is compact, then $f$ is Lipschitz on $X$. If $f$ is locally Lipschitz on $X$ and $X$ is compact, then $f$ is Lipschitz on $X$.
My proof: Since $f$ is locally Lipschitz on $X$, for each $x ∈ X$ there exists an
open $B_x$ containing $x$ such that $f$ is Lipschitz on $B_x$. Consider the collection
of all such $B_x$. This collection forms an open cover of $X$ and so there is a finite sub-collection $\{B_1, B_2, . . . , B_n\}$ which also covers $X$, since $X$ is compact. Since $f$ is Lipschitz on each $B_i$, there is an $B_i$ such that $d(f(x_i), f(y_i)) \leq M_i d(x_i, y_i)$ for all $x_i, y_i \in Wi$, for $i ∈ \{1, 2, . . . , n\}$. Then taking $B = max\{B_1, B_2, . . . , B_n\}$, we see that $f$ is Lipschitz on $X$.
Is my solution correct? 
Can we use the fact of compact metric space that every sequence in $X$ has a convergent sub-sequence to prove the same fact??

First note that locally Lipschitz implies continuous. This said, the argument suggested can be completed as follows. For every $x\in X$ there is an open ball $B_x$ centered at $x$ of radius $\varepsilon_x$ and a constant $M_x$ such that $d(f(x),f(y))\le M_xd(x,y)$ for $x,y\in B_x$. Now consider the open convering consisting of the open balls $B'_x$ of center $x$ and radius $\varepsilon_x/2$, and by compactness extract a finite subcovering $B'_{x_1},\dots,B'_{x_r}$. In this situation:
$$
\delta=\min\big\{\frac{\varepsilon_{x_i}}{2}: i=1,\dots,r\big\}>0\quad\text{and}\quad
K=\max\{d(f(x),f(y)): x,y\in X\} 
$$
exists by compactness and continuity. Then pick a positive constant
$$
L\ge \frac{K}{\delta}, M_{x_1},\dots,M_{x_r}.
$$
We see that $f$ is $L$-Lipschitz. Indeed, 
(1) if $d(x,y)<\delta$, as $x\in B'_{x_i}$ for some $i$ we have
$$
d(y,x_i)\le d(y,x)+d(x,x_i)<\delta+\frac{\varepsilon_{x_i}}{2}\le\varepsilon_{x_i},
$$
hence $x,y$ are both in $B_i$ and 
$$
d(f(x),f(y))<M_id(x,y)\le L\,d(x,y),
$$  
(2) if $d(x,y)\ge\delta$ we have
$$
d(f(x),f(y))\le K=\frac{K}{\delta}\delta\le L\,d(x,y).
$$
We are done.
