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\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 239/7:}
Let $p = (x_1,\dots,x_n), q = (y_1,\dots,y_n) \in E$ be any two points.
For $k=1,\dots, n$ let $p_k = (x_1,\dots,x_k,y_{k+1},\dots,y_n)$ (in
particular $p_n = p$), and set
$p_0 := 0$.  Then for $1\leq k\leq n$, the points $p_k$ and 
$p_{k-1}$ differ only in the $k$th coordinate.  Hence, by the one-variable
mean value theorem, there exists $c_k$ between $x_k$ and $y_k$ such that
$$
|f(p_k) - f(p_{k-1})| = |D_kf(c_k) (x_k - y_k)|
\leq C\norm{x_k - y_k},
$$
where $C>0$ is an upper bound for $D_1 f, \dots D_n f$ on $E$.  Hence
$$
|f(p) - f(q)| \leq \sum_{k=1}^n|f(p_k) - f(p_{k-1})|
\leq C\sum_{k=1}^n |x_k - y_k| \leq Cn\norm{p - q}.
$$
The main thing is that the constant $Cn$ has nothing to do with $p$ or $q$.

Now let $p\in E$ be any point and $\{p_n\}\subset E$ be any sequence convergin
to $p$.  Then 
$$
0 \leq \lim_{n\to\infty} \norm{f(p_n) - f(p)} 
\leq C\lim_{n\to\infty} \norm{p_n - p} = 0.
$$
In other words $\lim_{n\to\infty} f(p_n) = f(p)$, which shows that $f$ is
continuous at $p$.  Since $p$ was arbitrary, we conclude that $f$ is
continuous on $E$.
\qed
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