Proof

Let [latex]I[/latex] be a non-trivial interval on which [latex]f[/latex] is continuous and assume [latex]f[/latex] has no critical points in the interior of [latex]I[/latex].  Let [latex][a,b][/latex] be a non-trivial, closed subinterval of [latex]I[/latex].   By the extreme value theorem, [latex]f[/latex] assumes both a minimum value and a maximum value in this interval.  Neither can occur in the interior, as that would imply the existence of an
interior critical point.  Therefore, the minimum and maximum values occur at the
endpoints [latex]a[/latex] and [latex]b[/latex].  Moreover, these values cannot be equal, for then, [latex]f[/latex] would be
constant over the interval and all interior points would be critical.  It follows that [latex]f[/latex]
cannot take the same value at any two different points of [latex]I[/latex].

Assume that [latex]f[/latex] takes its minimum value at [latex]a[/latex] and its maximum value at [latex]b[/latex].  If this is the case, it remains to prove that [latex]f[/latex] is increasing over [latex]I[/latex].  If the reverse is true, a similar proof would show that [latex]f[/latex] is decreasing over [latex]I[/latex].

Claim:  [latex]f[/latex] is increasing on the interval [latex][a,b][/latex].
If not, there would exist values [latex]x_1[/latex] and [latex]x_2[/latex] in the interval with [latex]x_1 < x_2[/latex] and [latex]f(x_1) > f(x_2)[/latex].  Then, in the interval [latex][a,x_2][/latex], the minimum value would occur at the left endpoint [latex]a[/latex], but the maximum would not occur at the other endpoint.  As argued above, this is not possible.

It follows that [latex]f[/latex] is strictly monotone over any finite closed subinterval of [latex]I[/latex].

If [latex]I[/latex] itself is a finite closed interval, the proof is complete.

Otherwise, let [latex]x_1[/latex] and [latex]x_2[/latex] be any two points in [latex]I[/latex] with [latex]x_1 < x_2[/latex].  Select a finite closed subinterval [latex]F[/latex] of [latex]I[/latex] that contains these two points and also the original interval [latex][a,b][/latex].  Then [latex]f[/latex] is strictly monotone over [latex]F[/latex], and since it is increasing on [latex][a,b][/latex], it is also increasing on [latex]F[/latex].  Therefore, [latex]f(x_1) < f(x_2)[/latex], and it follows that [latex]f[/latex] is increasing on all of [latex]I[/latex].  This completes the proof.