Proving a Lipschitz inequality for continuously differentiable functions. So I'm trying to show that given $\phi:G\to\mathbb{R}^n$ where $G\subseteq\mathbb{R}^n$ continuously differentiable, that if $\sup_G\|D\phi\|<c$, then $\phi$ is Lipschitz with constant $c.$ I'm a little nervous about my proof, as my intuition for multivariable functions is very shaky. Could anyone verify if what I'm doing is legal, or will it get me sent to prison?
$\textbf{Proof:}$
To start we consider 
$$\Phi:[0,1]\to\mathbb{R}$$
$$t\mapsto\left<\phi(a+th)|\phi(a+h)-\phi(a)\right>.$$
For convenience we let
$\phi(a+h)-\phi(a)=(x_1,...,x_n).$
observe
$$\Phi'(t)=(\phi_1(a+th)x_1+...+\phi_n(a+th)x_n)'$$
$$=\left<\nabla\phi_1(a+th)|h\right>x_1+....+\left<\nabla\phi_n(a+th)|h\right>x_n$$
$$=\sum_{i,j=1}^n\partial_{x_i}\phi_j(a+th)h_ix_j=\left<D\phi(a+th)h|\phi(a+h)-\phi(a)\right>.$$
We compute
$$|\Phi(a+h)-\Phi(a)|=\left|\int_{0}^1\Phi'(t)\right|\leq\int_{0}^1|\Phi'(t)|=\int_0^1\left|\left<D\phi(a+th)h|\phi(a+h)-\phi(a)\right>\right|$$
$$\leq\int_{0}^1\|Df(a+th)h\|\|\phi(a+h)-\phi(a)\|\leq\int_0^1\|D\phi(a+th)\|\|h\|\|\phi(a+h)-\phi(a)\|$$
$$\leq c\|(a+h)-a\|\|\phi(a+h)-\phi(a)\|.\text{ }(*)$$
Next we observe that
$$|\Phi(a+h)-\Phi(a)|=\left<\phi(a+th)-\phi(a)|\phi(a+h)-\phi(a)\right>=\|\phi(a+h)-\phi(a)\|^2,$$
hence by $(*)$ we find
$$\|\phi(a+h)-\phi(a)\|\leq c\|(a+h)-a\|.$$
This shows that $\phi$ is Lipschitz with constant $c$ as desired. $\blacksquare$
 A: Not Guilty! Prison may be avoided in this case.
In fact, our OP Melody's actions in this matter appear, to this reviewer, to be pefectly legal.
These pronouncements having been made, perhaps I should say that I did at first find the calculations presented in the text of the question a little bit difficult to follow, especially the derivation of the formula
$\Phi'(t) = \left<D\phi(a+th)h|\phi(a+h)-\phi(a)\right>; \tag 0$
however, upon careful scrutiny I realized the maneuvers as yet another legitimate means to arrive at (0).  
So, lauds to our OP Melody.
My derivation of the requisite result is as follows:
Consider the line segment $\gamma(t)$ which joins $x$ and $y$:
$\gamma(t) = (1 - t)x + ty, \; t \in [0, 1]; \tag 1$
then
$\gamma'(t) = y - x; \tag 2$
we have
$\Vert \phi(y) - \phi(x) \Vert = \Vert \phi(\gamma(1)) - \phi(\gamma(0)) \Vert = \left \Vert \displaystyle \int_0^1 \dfrac{d\phi(\gamma(s))}{ds} \; ds  \right \Vert; \tag 3$
a simple application of the chain rule yields
$\dfrac{d\phi(\gamma(t))}{dt} = D\phi(\gamma(t))\gamma'(t) = D\phi(\gamma(t))(y - x), \tag 4$
which  may be substituted into (3), whence we find
$\Vert \phi(y) - \phi(x) \Vert = \left \Vert \displaystyle \int_0^1  D\phi(\gamma(s))(y - x) \; ds \right \Vert \le \displaystyle \int_0^1   \Vert D\phi(\gamma(s))(y - x) \Vert \; ds$
$\le \displaystyle \int_0^1 \Vert D\phi(\gamma(s)) \Vert \Vert (y - x) \Vert \; ds  \le \int_0^1 c \Vert (y - x) \Vert \; ds = c \Vert y - x \Vert \int_0^1 ds = c \Vert y - x \Vert, \tag 5$
or in short,
$\Vert \phi(y) - \phi(x) \Vert \le  c \Vert y - x \Vert, \tag 6$
and we see that $\phi(x)$ is indeed Lipschitz continuous with Lipschitz constant at most $c$.  $OE\Delta$.
Nota Bene:  Though not mentioned by our OP Melody, nor yet by Yours Truly, I should point out that we need to assume that $G$ is convex to ensure this demonstration flies, convexity here meaning that for any $x, y \in G$ the segment $\gamma(t)$ also lies in $G$:
$\forall t \in [0, 1], \; \gamma(t) = (1 - t)x + ty \in G; \tag 7$
this assumption guarantees that
$\Vert D\phi(\gamma(t)) \Vert \le c, \forall t \in [0, 1], \tag 8$
essential to the above proof that $\phi$ is Lipschitz.  End of Note.
