Equivalent characterization of differentiable quasi-concave functions I am looking for a proof for the following statement:

Suppose $f: A \to \mathbb R$, where $A$ is an open subset of $\mathbb R$, is continuously differentiable on $A$ which has the property that for all $x_2, x_1 \in A$,
$$f(x_2)\geq f(x_1) \implies (x_2-x_1)f'(x_1)\geq0$$
Prove that $f(x)$ is quasi-concave.

Attempt:
We need to show that for all $\theta \in [0,1]$, $f(\theta x_2 + (1-\theta)x_1) \geq f(x_1)$. For contradiction, suppose not. So, $\exists \,\,\bar \theta \in (0,1)$ such that $f(x_1) > f(\bar\theta x_2 + (1-\bar\theta)x_1)$.
Let $z\equiv \bar\theta x_2 + (1-\bar\theta)x_1$. From the property of $f$ stated above:
$$(x_1-z)f'(z) \geq 0, \quad (x_2-z)f'(z) \geq 0.$$
Since, $x_i\leq z\leq x_j$, $f'(z)=0$.
This is where I am stuck. I found some sources which use MVT hereafter to prove. But I didn't understand those proofs very well. I was trying to use converse of MVT but realized that it's wrong (i.e., I cannot claim that there exists some $z_1, z_2$ such that $z$ lies between them and $f(z_1) = f(z_2)$).
I want to be able to show somehow that $f'(z)=0$ is a contradiction to the property above.

EDIT: based on the posted answer, does this work:
Case1: $f(x_2)\geq f(x_1)$ and $x_2>x_1$
Let $x_n$ be an increasing sequence in $[x_1,x_2]$ such that $f(x_n) \geq f(x_1)$ for all $n$. Existence of such a sequence is without loss of generality as there will be finite or infinite points which satisfy $f(x_n) \geq f(x_1)$ and we are just taking an ordered sequence of them.
Since $[x_1,x_2]$ is bounded from above, this sequence will converge. Since the set is also a closed set, the sequence will converge in the set. Let $z$ be the limit point of this sequence. Clearly $z \neq x_2$. So there is an interval $(z,x_2]$ such that for all $x$ in this interval $f(x)<f(x_1)$. As discussed above and in the answer posted, $f'(x)$ for all $x$ in
$(z,x_2]$ is zero and thus the right limit of $f(z)$ is $f(x_2)$ which contradicts continuity.
 A: Recall the definition of quasiconcavity:$$
f((1 - t) x_1 + tx_2) \geqslant \min(f(x_1), f(x_2)),
$$
where $x_1, x_2 \in A$ and $t \in (0, 1)$. In fact, the condition can be further slightly weakened.

Proposition: If $f \in C([a, b])$ is differentiable on $(a, b)$, and for any $x_1, x_2 \in [a, b]$,\begin{gather*}
f(x_2) \geqslant f(x_1) \implies (x_2 - x_1) f'(x_1) \geqslant 0, \tag{1}
\end{gather*}
then $f$ is quasiconcave on $[a, b]$.

Proof: Suppose otherwise, then there exist $x_1, x_2 \in [a, b]$ with $x_1 < x_2$, and $t \in (0, 1)$ such that$$
f((1 - t) x_1 + tx_2) < \min(f(x_1), f(x_2)).
$$
Denote $x_3 = (1 - t) x_1 + tx_2$. Since $f$ is continuous at $x_3$, there exists $δ > 0$ with $f(x) < \min(f(x_1), f(x_2))$ for $x \in (x_3 - δ, x_3 + δ)$.
Case 1: $f(x_1) \leqslant f(x_2)$.
Define $y_1 = \sup\{x \in [x_1, x_3] \mid f(x) \geqslant f(x_1)\}$, then $y_1 \leqslant x_3 - δ$. On the one hand, for any $x \in (y_1, x_3)$, since$$
f(x) < f(x_1) = \min(f(x_1), f(x_2)),
$$
then (1) implies that$$
(x_1 - x) f'(x) \geqslant 0, \quad (x_2 - x) f'(x) \geqslant 0,
$$
thus $f'(x) = 0$. Therefore, $f$ is constant on $[y_1, x_3]$ by Lagrange's theorem, and $f(y_1) = f(x_3)$.
On the other hand, the definition of $y_1$ shows that there exists $\{u_n\} \subseteq [x_1, y_1)$ with $f(u_n) \geqslant f(x_1)$ for all $n$ and $u_n → y_1$ ($n → ∞$), hence the continuity of $f$ at $y_1$ implies that$$
f(y_1) = \lim_{n → ∞} f(u_n) \geqslant f(x_1) > f(x_3),
$$
a contradiction.
Case 2: $f(x_1) > f(x_2)$.
Define $y_2 = \inf\{x \in [x_3, x_2] \mid f(x) \geqslant f(x_2)\}$. The rest follows analogously to Case 1.

The appended discussion in the question has an error: For the selected sequence $\{x_n\}$, it can happen that $z = x_2$ as there is no other restriction.
