I would like some input in my attempted proof for this question:
Suppose that $f: \mathbb{R} \to \mathbb{R}$ is smooth and has at least two local minima. Show that $f$ has a critical point between these two minima.
$\textbf{My answer:}$
Let $x_{1}$ and $x_{2}$ be local minima of $f$. Without loss of generality, assume $x_{1} \lt x_{2}$. If $f(x_{1})=f(x_{2})$, it is trivial from the Mean Value Theorem that $\exists c \in (x_{1},x_{2}): f'(c)=0$.
Suppose $f(x_{1}) \lt f(x_{2})$ [the case where $f(x_{1}) \gt f(x_{2})$ envolves the same steps]. By the Mean Value Theorem, $\exists c \in (x_{1},x_{2}): f'(c) \gt 0$.
Since $x_{2}$ is a local minimum, $\exists \delta_{2} \gt 0:y \in (x_{2}-\delta_{2},x_{2}+\delta_{2}) \implies f(y) \geq f(x_{2})$.
Consider the closed interval $[x_{2}-\delta_{2},x_{2}]$. By the Mean Value Theorem, $\exists m \in (x_{2}-\delta_{2},x_{2}):f'(m)=\frac{f(x_{2})-f(x_{2}-\delta_{2})}{\delta_{2}} \lt 0$.
Since $f$ is smooth, $f'$ is continuous in $\mathbb{R}$. Therefore, we have $c,m \in (x_{1},x_{2})$ and $f'(c) \gt 0$ and $f'(m) \lt 0$, hence, $\exists z \in (c,m):f'(z)=0$.
Since the function is smooth we can use the first derivative as a continuous function. So, if $x_1<x_2$ are the two points of local minimum, we can say that, for $\epsilon_1, \epsilon_2 >0$ such that $x_1+\epsilon_2<x_2-\epsilon_2$ we have: $$ f'(x_1+\epsilon_1)>0 \quad \mbox{and}\quad f'(x_2-\epsilon_2)<0 $$ so, since $f'(x)$ is continuous there is a point $x_3$ in $[x_1+\epsilon_1,x_2-\epsilon_2]$ such that $f'(x_3)=0$
This seems essentially your idea, without the use of $f(x_1)$ and $f(x_2)$.