Alternative Proof of the Extreme Value Theorem I have proven the Boundedness Theorem for continuous functions and would now like to prove the Extreme Value Theorem; that is, show that the upper bound is indeed attained for continuous functions. I would like to use the most direct approach as possible, straight from the properties of real continuous functions (the approach in Spivak's Calculus seems too clever for me). I also want to avoid using sequences. 
Assuming that $f$ is continuous and bounded on $[a,b]$, here's what I've done so far:
Let $g(x) = \sup \{f(t), t \in [a,x] \}$ and $A = \{x : x \in [a,b] \text{ and } g(x)=g(b)\}$. Let $\alpha = \inf A$, then $\alpha \in [a,b]$, so $f(\alpha)$ is defined. Also $f(\alpha) \leq g(b)$ since $g(b) = \sup \{f(t) : t \in [a,b]\}$. So we only need to show that $f(\alpha) < g(b)$ entails a contradiction.
Assume $f(\alpha) < g(b)$, then $g(b)-f(\alpha) > 0$. Then by continuity of $f$ at $\alpha$, there is a $\delta$, such that $f(x)-f(\alpha) < g(b) - f(\alpha)$ and so $f(x) < g(b)$ for all $x \in [\alpha-\delta, \alpha + \delta]$. But since $\alpha$ is a greatest upper bound, there is an $x' \in [\alpha, \alpha + \delta]$ such that $x' \in A$ and so $g(x') = g(b)$.
Can this proof be finished or am I just headed for nowhere?
 A: Let's use the notation provided by OP. Thus we have $$g(x) = \sup\,\{f(t)\mid t \in [a, x]\},\,\, A = \inf\,\{x\mid x \in [a, b], g(x) = g(b)\},\,\, \alpha = \inf\, A$$ Clearly we can see that $M = g(b)$ is the supremum of $f$ on whole interval $[a, b]$ and $\alpha \in [a, b]$. And our objective is to find a point $x \in [a, b]$ such that $f(x) = M$.
If $f(a) = M$ then we are done. Hence let's assume that $f(a) < M$. This ensures that there is a range of values of $x$ near $a$ where $f(x) < M$ and hence we have an $h > 0$ such that $f(x) < (M + f(a))/2$ for all $x \in [a, a + h]$ (this is derived by using continuity of $f$ at $a$ using the value of $\epsilon = (M - f(a))/2$).  This implies that $g(x) < M$ for all $x \in [a, a + h]$. So $A$ does not contain any point of $[a, a + h]$ and hence $\alpha > a$. Same way we can assume that $f(b) \neq M$ and deduce that $\alpha < b$. Thus $\alpha \in (a, b)$.
We now claim that $f(\alpha) = M = g(b)$. Suppose that it is not the case then $f(\alpha) < M$ and let's put $\epsilon = (M - f(\alpha))/2$. Then we have a $\delta > 0$ such that $(\alpha - \delta, \alpha + \delta) \subseteq [a, b]$ and $$f(\alpha) - \epsilon < f(x) < f(\alpha) + \epsilon$$ for all $x \in (\alpha - \delta, \alpha + \delta)$. It follows that $$K = \sup\,\{f(x) \mid x \in (\alpha - \delta, \alpha + \delta)\}\leq f(\alpha) + \epsilon = \frac{M + f(\alpha)}{2} < M\tag{1}$$ Consider the number $\beta = \alpha -(\delta/2)$. Clearly $\alpha - \delta < \beta < \alpha$ and hence $\beta \notin A$. This means that $g(\beta) < g(b) = M$. Thus supremum of $f$ on interval $[a, \beta]$ is less than $M$ and from equation supremum of $f$ on $(\alpha - \delta, \alpha + \delta)$ is also less than $M$. Since these two intervals have an interior point in common it follows that supremum of $f$ on their union $[a, \alpha + \delta)$ is also less than $M$.
Now by definition of $\alpha = \inf\, A$ it follows that there is a $\gamma \in A$ such that $\alpha < \gamma < \alpha + \delta$ and hence supremum of $f$ on $[a, \gamma]$ is $g(b) = M$. This is contrary to the fact established in last paragraph (because $\gamma$ is interior point of $[a, a + \delta)$).
The contradiction shows that $f(\alpha) = M = g(b)$.
