Stuck on part (b) of Rising Sun Lemma exercise

Here's another full proof that uses the Intermediate Value Theorem. Probably not the one the book is intending, though.

Lemma: If $x \geq b$ then $f(x) \leq f(b)$. Proof: $b$ is not a shadow point.

Proof: Now, choose a value $c$ such that $f(b)<c<f(a)$.

Let $S = \{x \in [a,b] : f(x) = c\}$, which is closed as a preimage, and bounded. So $S$ is compact. It is also nonempty by IVT, so $x_0 = \max S$ exists.

I claim $x_0$ is not a shadow point. If it were, then there would be a $y > x_0$ such that $f(y) > c > f(b)$. By the lemma, $y < b$. So again applying the intermediate value theorem, there is a $z \in (y,b)$ such that $f(z) = c$. But then $z \in S$ and $z > \max S$, a contradiction.

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is not a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.