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?

Can we use the increment theorem for the proof ?
$\delta z= f_{x}\delta x+f_{y}\delta y+\epsilon_{1}\delta x+\epsilon_{2}\delta y$ where $\epsilon_{1}$ and$\epsilon_{2}$ tends to $0$ as $\delta x,\delta y$ tends to $0$.
But $\delta z =f(x+\delta x,y+\delta y)-f(x,y)$.  Applying boundedness of partial derivatives $f_{x}$ and $f_{y}$ we get as $\lim f(x+\delta x,y+\delta y)=f(x,y) $ as $\delta x,\delta y$ tends to $0$
