Proving $|f(y)-f(x)|\leq f(|y-x|)$ for $f(x)=-x\log_2(x)$ I need to prove (or disprove) the following claim: 
for $0 \leq x \leq y \leq 1$:
$$|f(y)-f(x)|\leq f(y-x)$$
where $f(x) = -x \log_2(x)$ if $x \in (0,1)$ and $f(x)=0$ elsewhere.
I managed to prove that: $f(y)-f(x) \leq f(y-x)$ using Lagrange's mean value theorem (MVT), here is my proof:
Note that $f'$ is decreasing for $x \geq 0$, also$f'(0) \rightarrow \infty$ when $x \rightarrow 0^+$.
Let $z = y-x$.
If $x \geq z$, then using MVT:
\begin{equation*} \begin{aligned}
f(y)-f(x) &= z f'(c_1) && \text{for  } c_1 \in (x, y) \\
f(0+z)-f(0) &= z f'(c_2) && \text{for  } c_2 \in (0, z)
\end{aligned} \end{equation*}
Since $f'$ is decreasing and $z \leq x$ we have that: 
$$ f(y)-f(x) = z f'(c_1) \leq z f'(c_2) = f(0+z)-f(0) = f(z) $$
If $x < z$, again, using MVT:
\begin{equation*} \begin{aligned}
f(x)-f(0) &= x f'(c_1) && \text{for  } c_1 \in (0, x) \\
f(y)-f(z) &= x f'(c_2) && \text{for  } c_2 \in (z, y)
\end{aligned} \end{equation*}
Since $f'$ is decreasing and $x < z$ we have that: $x f'(c_2) \leq x f'(c_1)$ which implies:
\begin{align*}
f(y)-f(x) 
&= (f(y)-f(z)) + (f(z)-f(x)) \\
&= x f'(c_2) + (f(z)-f(x)) \\
&\leq x f'(c_1) + (f(z)-f(x)) \\
&= (f(x)-f(0)) + (f(z)-f(x)) \\
&= f(z)
\end{align*}
Now, I'm stuck here... Any idea of how to prove or disprove the claim with absolute value?

The claim is false as stated. Indeed, consider the case when $y=1$ and $x=\frac{1}{4}$. Then
$$
f(y) =f(1)= 0 \qquad\text{and}\qquad
f(x) =f\left(\frac{1}{4}\right)=\frac{1}{4}\log(4)
$$
and 
$$
f(y-x) = f\left(\frac{3}{4}\right) = \frac{3}{4}\log\left(\frac{4}{3}\right) = \frac{1}{4}\log\left(\frac{4^3}{3^3}\right).
$$
Note that $4>\frac{4^3}{3^3}$ and thus
\begin{align*}
\left\lvert f(y)-f(x)\right\rvert &= \frac{1}{4}\log(4) \\
                                  &> \frac{1}{4}\log\left(\frac{4^3}{3^3}\right)\\
& = f(y-x).
\end{align*}
Other counterexamples with $y\neq1$ can be generated and checked numerically. For example,  if we choose $x=0.2$ and $y=0.99$, then $f(y−x)−|f(y)−f(x)|≈−0.125$.
The claim is true if we remove the absolte value. Note that $f$ is concave, since $f''(x)=-\frac{1}{x}$. The claim that $f(y)-f(x)\leq f(y-x)$ follows from concavity of $f$, since
$$
f(0)+f(y)\leq f(x) +f(y-x)
$$
and $f(0)=0$.
