Prove that a continuous function on a closed interval attains a maximum As the title indicates, I'd like to prove the following:

If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.

Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.
What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
 A: Yes, quite a few such proofs exist. You first prove that a continuous function is bounded, and apply the Bolzano-Weierstrass theorem. Here’s such a proof from my lecture notes in a first course in analysis:

By the earlier result, $\,f$ is bounded. Consider the image of $[a,b]$ under $f$: $$A = f([a,b]) = \left\{f(x) : x \in [a,b]\right\}.$$
  $f$ bounded means that $A$ is bounded as a subset of $\mathbb{R}$. We also have that $A\neq \emptyset$. We will prove the existence of $x_2$, and the existence of $x_1$ is proved similarly.
Since $A$ is bounded and non-empty, there is a supremum: $M=\sup A$. Given any positive integer $n$, $M-1/n$ cannot be an upper bound for $A$. Then there exists some $x_n \in [a,b]$ such that %
  $$M-\frac{1}{n} < f(x_n) \leq M. \tag{$*$}$$
  By Bolzano-Weierstrass, $x_n$ has a convergent subsequence $x_{n_{j}} \to x$. Since $a\leq x_{n_{j}}\leq b$, $a\leq x\leq b$. By the continuity of $f$, $\,f(x_{n_{j}}) \to f(x)$. But by $(*)$, $\,f(x_{n_{j}}) \to M$. By the uniqueness of the limit, $\,f(x)=M$. Now set $x_2=x$. $\square$

A: Here’s a sketch of one possible argument. Let $u=\sup_{x\in[a,b]}f(x)$. (Note that I allow the possibility that $u=\infty$.) There is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $[a,b]$ such that for each $n\in\Bbb N$, $u-f(x_n)<\frac1{2^n}$ if $u\in\Bbb R$ and $f(x_n)>n$ if $u=\infty$. Extract a monotone subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$. Being a monotone, bounded sequence, $\langle x_{n_k}:k\in\Bbb N\rangle$ converges to some $y$. (Note that you have to use the completeness of $\Bbb R$ in some way, and this is the most elementary that occurs to me.) Moreover, $y\in[a,b]$, and $f$ is continuous, so $f(y)=\lim\limits_{k\to\infty}f(x_{n_k})=u$.
A: This theorem ultimately depends on the completeness of real number system and can't be proven without using any result equivalent to the completeness of real number system. If we assume that every continuous function on a closed interval is bounded then we can provide a very simple proof of the current problem being discussed here.
Under this assumption we have the existence of $M = \sup\{f(x)\mid x \in [a, b]\}$. If $f(x)\neq M$ for all $x \in [a, b]$ then the function $g(x) = 1/(M - f(x))$ is continuous in $[a, b]$ and hence bounded in $[a, b]$. On the other hand by definition of $M$ we can find $x \in [a, b]$ such that $M - f(x)$ can be made arbitrarily small. Thus $g(x)$ can be made arbitrarily large. This contradiction shows that we must have $f(x) = M$ for some $x \in [a, b]$.
The result that every continuous function is bounded on a closed interval is itself another property of continuous functions which can't be proved without using completeness of real number system.
I have presented various proofs of these properties of continuous function here.
A: I don't know how you can avoid compactness entirely. Here's one that uses sequential compactness in the form of the Bolzano-Weierstrass theorem, "every bounded sequence has a convergent subsequence". (To justify that theorem to beginners, you could take it as an axiom that every bounded monotonic sequence converges, and show them the proof that every sequence has a monotonic subsequence.) For some reason that escapes me, this proof is often done in two stages, first proving that the function is bounded and then showing that the it attains a maximum.
Given a continuous real-valued function $f$ on $[a,b]$, we will show that
the set $Y = f([a,b])$ has a greatest element.
For each positive integer $n$, define a finite set $Q_n = \{p/q: p,q \text{ integers, } 0 < q \le n, |p| \le n\}$.
Choose $y_n\in Y$ so as to maximize the number of elements in the set $\{r\in Q_n: y_n > r\}$, and choose $x_n\in[a,b]$ with $f(x_n) = y_n$.
The sequence $\{x_n\}$ has a subsequence converging to a point $c\in[a,b]$. Since $f$ is continuous, the corresponding subsequence of $\{y_n\}$ converges to $f(c)$. We will show that $f(c)$ is the greatest element of $Y$.
Assume for a contradiction that $f(c)<y\in Y$. Choose a rational
number $r$ so that $f(c)<r<y$. Because of the way $y_n$ was chosen, we
have $y_n > r$ whenever $r\in Q_n$. Since $r\in Q_n$ for all
sufficiently large $n$, we have $y_n > r > f(c)$ for all sufficiently
large $n$. But this is absurd, since $\{y_n\}$ has a subsequence converging
to $f(c)$.
A: I found the following proof by Takeshi Saito in "How to Learn Mathematics(New Edition)" edited by Kunihiko Kodaira.
Let $X$ be a non-empty compact space.
Let $f(x)$ be a real valued continuous function on $X$.
Assume that $f(x)$ doesn't attain a maximum value on $X$.
Then for any $x \in X$, there exists $t \in X$ such that $f(x) < f(t)$.
Let $U_t := \{x\in X | f(x) < f(t)\}$ for $t \in X$.
$f(x)$ is continuous on $X$, so $U_t$ is an open set of $X$.
$f(x)$ doesn't attain a maximum value on $X$, so $(U_t)_{t\in X}$ is an open cover of $X$.
$X$ is compact, so there exists a finite subcover $U_{t_1}, \cdots, U_{t_n}$.
Let $k \in \{1, \cdots, n\}$ be an integer such that $f(t_k) = \max \{f(t_1), \cdots, f(t_n)\}$.
$t_k \in X$. So there exists an integer $l$ such that $t_k \in U_{t_l}$.
Then, $f(t_k) < f(t_l) \leq f(t_k)$. This is a contradiction.
So, $f(x)$ attains a maximum value on $X$.
We now prove that $[0, 1]$ is compact.
The discrete space $\{0, 1\}$ is finite. So $\{0, 1\}$ is compact. So, $\{0,1\}^{\mathbb{N}}$ is compact.
Let $f : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$ be a function such that $f(\{a_n\}) := \sum_{n=1}^\infty \frac{a_n}{2^n}$.
Then, $f$ is continuous, so $f(\{0,1\}^{\mathbb{N}}) = [0,1]$ is compact.
Let $a, b \in \mathbb{R}$ such that $a \leq b$.
Then, $[a, b] \ne \emptyset$.
And it is easy to show that $[a, b]$ is a compact space:
Let $g : [0, 1] \to [a, b]$ be a function such that $g(x) = (b - a) x + a$.
Since $g$ is continuous, $g([0, 1]) = [a, b]$ is compact.
So, if $f(x)$ is a real valued continuous function on $[a, b]$, then $f(x)$ attains a maximum value on $[a, b]$.
A: Theorem. Let $f:[a,b]\to\mathbb{R}$ be continuous. Then $f([a,b])$ is bounded. 
Proof (indirectly) avoiding compactness: Since $f$ is continuous at $a,$ there exists $\delta>0$ such that for any $x\in [a,a+\delta),$ we have $f(a)-1< f(x)<f(a)+1.$ Hence $f$ is bounded on $[a,a+\delta/2].$ Write 
$$A=\{x\in [a,b]: f \mbox{ is bounded on } [a,x]\}.$$ 
Then $a+\delta/2\in A,$ and $b$ is an upper bound of $A.$ So, let $c=\sup(A).$  Now, $a+\delta/2 \leq c\leq b.$ Continuity of $f$ at $c$ implies that there exists $\delta_1>0$ such that for any $x\in (c-\delta_1,c+\delta_1)\cap [a,b],$ we have $f(c)-1<f(x)<f(c)+1.$ That is, $f$ is bounded on $(c-\delta_1,c+\delta_1)\cap [a,b].$ Let $d\in (c-\delta_1,c).$ Then $f$ is bounded on $[a,d].$ It follows that 
$$f \mbox{ is bounded on } [a,d]\cup\big( (c-\delta_1,c+\delta_1)\cap [a,b]\big).$$
Since $[a,c]\subseteq [a,d]\cup\big( (c-\delta_1,c+\delta_1)\cap [a,b]\big),$ we have $c\in A.$ 
Further, if $c<b,$ then choose $\delta_2=\min\{\delta_1,(b-c)/2\}.$ Then $[a,c+\delta_2]\subseteq [a,d]\cup\big( (c-\delta_1,c+\delta_1)\cap [a,b]\big).$ It follows that $f$ is bounded on $[a,c+\delta_2].$ So, $c+\delta_2\in A.$ This contradicts the fact that $c=\sup(A).$ Therefore, $c=b.$ 
Hence $c=b$ and $c\in A.$ That is, $b\in A.$ Therefore, $f$ is bonded on $[a,b].$   QED 
