The proof looks pretty straightforward, but with a few hiccups. Two, to be precise, which are restated at the bottom.
Th. For open set $O$, if function $F:O\rightarrow\mathbb R^n$ is $C^1$, then $F$ is locally Lipschitz.
Given Proof: the proof examines the compact ball $O_\epsilon$ of radius $\epsilon$ about arbitrary point $X_0$, and asserts that $O_\epsilon$ is convex, allowing a line segment between points $Y,Z\in O_\epsilon,$ given by $Y+sU$, where $U=Z-Y$, and $s\in [0,1]$.
Letting $\psi(s) = F(Y+sU)$, the proof proceeds to take the derivative, to find $$\psi'(s) = DF_{Y+sU}(U)$$ Where $DF_X(\cdot)$ is the book's notation for the $n\times n$ Jacobian of $F$ wrt vector $X$. But the choice of notation with $Y+sU$ in the subscript is throwing me for a loop: I'm unsure how to interpret that Jacobian. (Question 1: how should I understand $DF_{Y+sU}(U)$)
The proof continues, Therefore, $$F(Z)-F(Y) - \psi(1)-\psi(0)$$ $$= \int^1_0\psi'(s)ds$$ $$= \int^1_0DF_{Y+sU}(U)ds$$ Thus, $$|F(Z) - F(Y)| \le \int^1_0 K|U|ds = K|Z-Y|$$ Thereby satisfying definition of the Lipschitz condition. But, Question 2: I don't follow why that inequality should hold? That seems wholly unexplained.
Question 1: how should I understand $DF_{Y+sU}(U)$?
Question 2: How does the final inequality hold?
$DF_{Y+sU}(U)$ is the Jacobian matrix of $F$ computed at $Y+sU$ and then applied to the vector $U$. For example, for $n=1$ you have $$ \psi'(s) = F'(Y+sU)U, $$ using the chain rule.
For the second you have $$ |DF_{Y+sU}(U)|\le \|DF_{Y+sU}\|\cdot|U|\le K |U|, $$ where $K$ is an upper bound for the continuous function $s\mapsto \|DF_{Y+sU}\|$ on the compact set $[0,1]$.