Without use of Darboux's theorem, prove that $f'$, where $f(x)=x^2\sin\left(\frac{1}{x}\right)$, enjoys IVP 
Prove (wihout use of Darboux's theorem) that the derivative of the function: $$f(x)=\left \{\begin {array}{ll}
x^2\sin\left(\frac{1}{x}\right)&,~x\neq0\\
0&,~x=0\\
\end{array}
\right.,$$ that is
  $$f'(x)=\left \{\begin {array}{ll}
2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right) &\textrm{,  $x \neq 0$}\\
0 &\textrm{,  $x =0$}
\end{array}
\right.$$
  (see the pic), enjoys the Intermediate Value Property (IVP).

This function ($f'$) is a classic counterexample that IVP does not characterize continuity (it is not continuous at $0$). All the proofs I have seen argue on IVP using Darboux's theorem on derivative. 
Is there a way to give a straightforward proof in this particular function $f'$?
(one of course may restrict on intervals $I$ containing the discontinuity $0$, since $f'$ is continuous on every interval $I$ not containing $0$, so by IVT for continuous functions we get the desired result).
Thanks in advance.
 A: One keeps all definitions as in the question.
Lemma. Let $I$ be an interval either of the form $(\alpha ,0]$ or $[0,\beta)$. Let $J:=I\setminus \{0\}$. The $f'(J)$ contains the interval $(-1,1)$.
Proof. It is clear that $$\limsup_{x\rightarrow 0^+}f'(x)=1,\liminf_{x\rightarrow 0^+}f'(x)=-1,$$ and similarly
$$\limsup_{x\rightarrow 0^-}f'(x)=1,\liminf_{x\rightarrow 0^-}f'(x)=-1.$$
One prove the case when $J=(0,\beta)$ (the other case being similar). If $y_0\in (-1,1)$, then $y_0\in (-1+\epsilon,1-\epsilon)$ for som positive $\epsilon$.  From the above observations, there exists $x_2\in J$ such that $f'(x_2)>1-\epsilon$. And then there exists $0<x_1<x_2$ such that $f'(x_1)<-1+\epsilon$. Now since $(x_1,x_2)\subseteq J$ and $f'$ is continuous there, there exists by IVP for $f'$ that $\exists x_0\in (x_1,x_2)$ with $f'(x_0)=y_0.$ This completes the proof.
Proposition. $f'$ satisfies the IVP, namely for $I=[a,b]$, $f'$ assumes any value $y$ between $f'(a)$ and $f'(b)$.
Proof. As the OP remarked, it suffices to prove the case when $0\in I$ (so $a\leq 0,b\geq 0$). Also the case when $f'(a)=f'(b)$ being void, one assumes that $f'(a)\neq f'(b).$ Let $y$ be any value between $f'(a)$ and $f'(b)$. Now one may divide into two cases.
Case 1. $|y|<1.$
This follows from the lemma above.
Case 2. There are $4$ subcases (possibly overlapping) as follows (where one applies the lemma above to choose $x_0$):
Subcase 1. $1\leq y<f'(b)$: Take $0<x_0<b$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[x_0,b]$.
Subcase 2. $1\leq y<f'(a)$: Take $a<x_0<0$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 3. $f'(a) < y\leq -1$: Take $a<x_0<0$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 4. $f'(b)<y\leq -1$: Take $0<x_0<b$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[x_0,b]$.
Combining all cases, the proposition is proven.
