I wonder if this proof is alright and I can take it into my notebook. It is my version of the standard version, so that it makes more sense to me. If there are significant mistakes or shortcomings, please give me advice or corrections.
Theorem:
Let $f: [a,b] \to \mathbb{R}$ be a continuous function with $f(a) \leq f(b)$. Then for every $c \in [f(a),f(b)]$, there exists an $r \in [a,b]$ such that $f(r) = c$.
Proof:
Assume $f(a) < c < f(b)$, because if $c = f(a)$ or $c = f(b)$, then $r = a$ or $r = b$ respectively, and if $f(a) = f(b) = c$, then $r = a = b$, which means the above restriction is not problematic.
We define D := {$x \in [a,b] \mid f(x) < c$}. $D$ is not empty, because $f(a) < c$ and thus $a \in D$. Furthermore, $D \subseteq [a,b]$, so $D$ is bounded, and by the completeness axiom, $D$ has a supremum, denoted by $r$.
Given the total order of real numbers, there are exactly three possibilities for the function value at $r$: $f(r) > c$, $f(r) < c$, or $f(r) = c$.
(1) $f(r) > c$:
In this case, $r \notin D$, so $D$ must be infinite, because if $D$ were finite, the least upper bound (supremum) would simply be the largest element in $D$, and thus $r$ would be in $D$. Since $r$ is the supremum of $D$, there must be at least one sequence $x_n$ in $D$ with $r$ as its limit. Because $f$ is continuous, the sequence of function values $f(x_n)$ also converges to the function value at $r$. But since $x_n \in D$, $f(x_n) < c$, and therefore $f(r) \leq c$, because otherwise, some elements of the sequence $f(x_n)$ would have to satisfy $f(x_n) > c$. This contradicts the assumption in case (1).
(2) $f(r) < c$:
Here, $r \in D$ and $r < b$, because if $r = b$, then $f(r) = f(b) < c$, contradicting the initial premise. Since $r$ is the supremum of $D$ (in other words, the last element in $D$ whose function value is still less than $c$), and $f(b) > c$, it follows that for all $x \in (r,b]$, $f(x) \geq c$. Now, consider a sequence $x_n$ from $(r,b]$ converging to $r$. Because $f$ is continuous, the sequence of function values $f(x_n)$ converges to $f(r)$. But since $f(x) \geq c$, it follows that $f(x_n) \geq c$, and thus $f(r) \geq c$, because otherwise, some elements of the sequence $f(x_n)$ would have to satisfy $f(x_n) < c$, implying $f(x) < c$. This contradicts the assumption in case (2).
From this, it immediately follows that:
(3) $f(r) = c$:
We can apply this reasoning for any $c$ with $f(a) \leq c \leq f(b)$, and it will always hold that $f(r) = c$ with $r \in [a,b]$, though the specific values for $c$, $r$, and the content of the set $D$ would vary.
