= Solution
For the <entropy summand> $f(x)=-x\log x$, differentiation on $(0,1)$ gives
$$
f'(x)=-\frac{\ln x+1}{\ln2},\qquad f''(x)=-\frac1{x\ln2}<0.
$$
Since $f(x)\to0$ as $x\downarrow0$, its continuous extension to $[0,1]$ is a <concave function>.
Fix $0<v<1$ and define the increment $D_v(t)=f(t+v)-f(t)$ on $[0,1-v]$. In its interior,
$$
D_v'(t)=f'(t+v)-f'(t)=-\log\frac{t+v}{t}<0.
$$
Thus the increment is decreasing, and its endpoint values are $D_v(0)=f(v)$ and $D_v(1-v)=-f(1-v)$. Therefore
$$
-f(1-v)\leq f(t+v)-f(t)\leq f(v),
\qquad
\boxed{|f(t+v)-f(t)|\leq\max\{f(v),f(1-v)\}.}
$$
The cases $v=0,1$ follow directly from the endpoint values.
To obtain the sharper half-interval bound, put $G(v)=f(v)-f(1-v)$. It has $G(0)=G(1/2)=0$, while
$$
G''(v)=\frac{2v-1}{v(1-v)\ln2}\leq0\qquad(0<v\leq1/2).
$$
Hence $G$ is a <concave function> on $[0,1/2]$ and lies above the chord joining its zero endpoints. Thus $f(v)\geq f(1-v)$ there, proving
$$
\boxed{|f(t+v)-f(t)|\leq f(v)\qquad(0\leq v\leq1/2).}
$$
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