If $f$ is differentiable on [a,b], must there be a subinterval on which$f’(x)$ be bounded? If $f(x)$ is differentiable on $[a,b]$, then there must be $(\alpha,\beta)\subset [a,b]$, such that $f’(x)$ is bounded on $(\alpha,\beta)$. To prove it or to give an counter-example.
The question may have something to do with the intermediate value theorem.
I understand if there is no such $(\alpha,\beta)$, then for every $x\in[a,b]$, there exists ${x_n}$ such that $\lim_{n\rightarrow\infty} x_n=x$, and $\lim_{n\rightarrow\infty} |f’(x_n)|=+\infty$. But I still cannot work things out.
 A: $f'(x)=\lim \frac {f(x+\frac  1n) -f(x)} {1/n}$. It is a known consequence of Baire Category Theorem that pointwise limits of continuous functions have lots of points of continuity: They are continuous on a dense set of points. If $f'$ is continuous at $c$ then it is bounded in some interval around $c$.
A: Every derivative is bounded on a dense set of intervals.
(StackExchange says to be nice to a new contributor. Hence the elaboration on the other [correct] posted answer.)
The current answer is what any trained mathematician will respond.  If you wake us at 3:00am after a night of drinking we may not remeber much about the evening but if you ask this question we will mutter immediately (exactly as did the other poster):
(i)  Every derivative belongs to the first class of Baire. (See ref [1].)
(ii) Every function in the first class of Baire is continuous at the points of a residual set.  (See ref [2].)
(iii) If a function is continuous at a point it is necessarily bounded in a neighborhood of that point. (See any calculus text.)
From those three observations  the answer is clear: a derivative must be bounded on each of a dense set of subintervals.  The only "deep" step is (ii) and that is proved using the Baire category theorem.

ALTERNATIVELY: The OP suggested that, maybe, some more elementary property of derivatives and continuous functions might answer this, e.g., the mean-value theorem perhaps?
My guess is that you need the Baire category theorem to get something like this, although an elementary method is worth a try.  Arguably, though, the Baire category theorem is not that deep and should be a ready tool for a student anyway.
In the text [3] see Section 6.4 The Baire Category Theorem for an account of this theorem which is not particularly demanding or advanced.  Many students may encounter this theorem only in a study of complete metric spaces.
CONSIDER THIS METHOD:
If $f:[a,b]\to \mathbb R$ is continuous and has a finite derivative at each point of a set $D$ then, for each $n=1,2,3,\dots$, let  $E_n$ be the set of points $x$ with the property that
$$ y\in [a,b],\ 0<|x-y|\leq \frac1n \implies \left|\frac{f(y)-f(x)}{y-x}\right|\leq n.$$
Check (using continuity) that each $E_n$ is closed and that every point in $D$ belongs to at least one of the sets in the sequence $\{E_n\}$.
Now bring out the heavy guns:  if $D=[a,b]$ the sequence of closed sets $\{E_n\}$ covers that interval so, by the Baire category theorem one of the sets must contain an interval  $(c,d)$.  On that interval $f'$ is evidently bounded.
Note: the other method is not about derivatives, really only about pointwise limits of continuous functions.  This technique is useful for a detailed analysis of derivatives in more general situations.

REFERENCES:
[1] About references for Baire class one function and Baire class two function
[2] https://mathoverflow.net/questions/32033/points-of-continuity-of-baire-class-one-functions
[3] Elementary Real Analysis text (free PDF):
https://classicalrealanalysis.info/documents/TBB-AllChapters-Landscape.pdf
