About a Lipschitz condition. In his book Differential Manifolds, Lang states the following:

If $f : U \subseteq E \rightarrow E~$ is a $C^1$ map, where $E$ is a Banach space and
   $U$ is an open set, then it follows at once from the Mean Value Theorem that $f$ satisfies a Lipschitz condition on an open neighborhood $U_0$ of any given point $x_0  \in U$.

I don't know why this should be true.
I think that the proof would normally proceed along these lines: fix $x_0 \in U$, take the supremum
$$K = \sup_{x \in B} \|Df(x)\|,$$
where $B$ is some small closed ball around $x_0$, and that's your Lipschitz constant. The only problem is that since $E$ might be infinite-dimensional, I don't see why the supremum above should be finite.
If this statement is incorrect, can you provide a specific counterexample?
Thanks!

Why not? By continuity (we are consider $C^1$ functions) the mapping 
$$ x \mapsto \|Df(x)\| $$
as a mapping $U \to \mathbb{R}$ is continuous. By definition this means that for every $x$ and every $\epsilon > 0$ there exists an open neighborhood $U_\epsilon$ such that $x\in U_\epsilon$ and that for every $y\in U_\epsilon$, $|\|Df(x)\| - \|Df(y)\|| < \epsilon$. This in particular tells you that in a small open neighborhood of any point, the values $\|Df\|$ remain bounded. 

You seem to think that compactness is used here somewhere (I based this guess on your statement of finite dimensionality). It is not. The supremum will exist (and is bounded by continuity); it just may not be attained. 
