Bounded partial derivatives imply continuity As stated in my notes:

Remark: Suppose $f: E \to \mathbb{R}$, $E \subseteq \mathbb{R}^n$, and $p \in E$.  Also, suppose that $D_if$ exists in some neighborhood of $p$, say, $N(p, h)$ where $h>0$.  If all partial derivatives of $f$ are bounded, then $f$ is continuous on $E$.

I found a sketch of the proof here.  I'm wondering if I can adapt this proof as follows:
$f(x_1+h_1,...,x_n+h_n)-f(x_1,...,x_n)=f(x_1+h_1,...,x_n+h_n)-f(x_1,x_2+h_2,...,x_n+h_n)-...-f(x_1,x_2,...,x_{n-1}+h_{n-1},x_n+h_n)-f(x_1,...,x_{n-1},x_n+h_n)-f(x_1,...,x_n)$
However, I'm not sure how to apply the contraction principle to finish off the proof.  Is there a more efficient way to prove the above remark?

The proof is a combination of two facts:


*

*A function of one real variable with a bounded derivative is Lipschitz. 

*Let $Q\subset \mathbb R^n$ be a cube aligned to coordinate axes. If  a function $f:Q\to\mathbb R$ is Lipschitz in each variable separately, then it is Lipschitz. 
The proof of 2 involves a telescoping sum such as 
$$\begin{split}
f(x,y,z)-f(x',y',z')&= f(x,y,z)-f(x',y,z) \\ & + f(x',y,z)-f(x',y',z)\\&+f(x',y',z)-f(x',y',z')
\end{split}$$
