#### Answer

Interval: $[\frac{9}{2},\ 5)\cup (5,\ \frac{11}{2}]$

#### Work Step by Step

$0\lt|x-5| \leq\frac{1}{2}$
$\frac{1}{2}\leq x-5\leq\frac{1}{2}$
$\frac{1}{2}+5\leq x-5+5\leq\frac{1}{2}+5$
$\frac{9}{2}\leq x\leq\frac{11}{2}$
$4.5\leq x\leq5.5$
However, we exclude $x=5$ because then we would have $0\lt0$, which is false.
Interval: $[\frac{9}{2},\ 5)\cup (5,\ \frac{11}{2}]$