Let $f:\mathbb R^d \to \mathbb R^m$ be a map of class $C^1$. That is, $f$ is continuous and its derivative exists and is also continuous. Why is $f$ locally Lipschitz?
Such $f$ will not be globally Lipschitz in general, as the one-dimensional example $f(x)=x^2$ shows: for this example, $|f(x+1)-f(x)| = |2x+1|$ is unbounded.
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If $f:\Omega\to{\mathbb R}^m$ is continuously differentiable on the open set $\Omega\subset{\mathbb R}^d$, then for each point $p\in\Omega$ there is a convex neighborhood $U$ of $p$ such that all partial derivatives $f_{i.k}:={\partial f_i\over \partial x_k}$ are bounded by some constant $M>0$ in $U$. Using Schwarz' inequality one then easily proves that $$\|df(x)\|\ \leq\sqrt{dm}\>M=:L$$ for all $x\in U$. Now let $a$, $b$ be two arbitrary points in $U$ and consider the auxiliary function $$\phi(t):=f\bigl(a+t(b-a)\bigr)\qquad(0\leq t\leq1)$$ which computes the values of $f$ along the segment connecting $a$ and $b$. By means of the chain rule we obtain $$f(b)-f(a)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=\int_0^1df\bigl(a+t(b-a)\bigr).(b-a)\>dt\ .$$ Since all points $a+t(b-a)$ lie in $U$ one has $$\bigl|df\bigl(a+t(b-a)\bigr).(b-a)\bigr|\leq L\>|b-a|\qquad(0\leq t\leq1)\>;$$ therefore we get $$|f(b)-f(a)|\leq L\>|b-a|\ .$$ This proves that $f$ is Lipschitz-continuous in $U$ with Lipschitz constant $L$.
Maybe this can help. The Lipschitz condition comes many times from the Mean Value Theorem. Search the link for the multivariable case. The fact that $f$ is $C^1$ helps you to see that when restricted to a compact set the differential is bounded. That's why you only have local Lipschitz condition.