Theorem: Let $f$ be a continuous real-valued function on a closed rectangle $R$ in $\Bbb R^2$. Then,
(a) $\quad f$ is bounded on $R$
(b) $\quad $There exist points $c$ and $d$ in $R$ so that $f(c) = \sup_{p\in R} f(p), \quad f(d) = \inf_{p\in R} f(p)$.
(c) $\quad f$ is uniformly continuous on $R$.
My Work
(a) $\quad$ To prove that $f$ is bounded on $R$, we need to show that there is an $U$ so that $|f(p)| \le U$ for all $p \in R$. Suppose this is false. Then for each large integer $n$, there is a $p_n \in R$ such that $|f(p_n)| > n$. By Bolzano-Weierstrass, the bounded sequence $\{p_n\}$ has a convergent subsequence $\{p_{n_k}\}$ that converges to $p_0$ as $k \to \infty$. Since $f$ continuous, $f(p_{n_k}) \to f(p_0)$ as $k \to \infty$. But this is impossible since $f(p_{n_k})$ is unbounded by virtue of $f(p_n)$ being unbounded. So $f$ is bounded on $R$.
(b) $\quad$ Since $S = \{f(x) : x \in R \}$ is nonempty, $S$ has a supremum. Call this supremum $M$. Consider a sequence $\{c_n\}$ and the interval $M – \frac{1}{n} \le f(c_n) \le M$. Since this interval is closed, $f$ must have a convergent subsequence $\{c_{n_k}\}$ that converges as $k \to \infty$. Taking $N \ge 1/\epsilon$, we have $M – (M – 1/n) = 1/n \le \epsilon$ for $n \ge N$. So since $f$ is continuous we must have $f(c_n) \to f(c) = M$.
My Question: I feel pretty sure of my proof of (a), but (b) seems like it lacks a key component. And I am not sure how to prove (c) for any arbitrary $f$, given that the definition of uniform continuity for a continuous $f$ on a set $E \subseteq \Bbb R ^2$ is given $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \|q – p \| \le \delta \ \Rightarrow \ |f(q) – f(p) | \le \epsilon \quad \forall \ q, p \in E $$
Edit
Revised part (c):
If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have
$$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. So for some large integer $K \ge 1/\epsilon$, and $k \ge K$, we have $\Vert y_{n_k} – x \Vert \le \epsilon$ so $(y_{n_k})$ converges to $x$ as well.
From this we have that $(x_{n_k}), (y_{n_k}) \to x$, but $\Vert f(x_{n_k}) – f(y_{n_k}) \Vert \ge \epsilon_0$ which implies that $f$ is not continuous at $x$ which contradicts the hypothesis.
Best Answer
For part b), you cannot start by saying "$f$ achieves a maximum...". This is what you are trying to prove. Instead, start by observing that the set $\{f(x)|x\in R\}$ is nonempty and bounded above. It thus has a supremum, $U$. Now (carefully) define your $c_n$ so that $U-1/n\le f(c_n)\le U$ and continue with your argument.
For part c), you could argue as follows:
If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have $$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. Show that the corresponding subsequence $(y_{n_k})$ of $(y_n)$ also converges to $x$. Now, show that this and $(1)$ imply that $f$ is not continuous at $x$.