= Solution
Put $u=\varepsilon/2$. The state is well defined for $0\leq\varepsilon\leq2$, with eigenvalues $1-u,u$. The first state is pure and has zero <Von Neumann entropy>, so the exact difference is
$$
\boxed{|S(\rho)-S(\sigma)|=h_2(\varepsilon/2),\qquad
h_2(u)=-u\log_2u-(1-u)\log_2(1-u).}
$$
This is the <entropy of a binary diagonal qubit mixture>. Its <trace distance> from the <pure state> is $u$, not $\varepsilon$. An explicit elementary upper bound follows from $-(1-u)\ln(1-u)\leq u$:
$$
\boxed{|S(\rho)-S(\sigma)|\leq
\min\left\{1,\frac{\varepsilon}{2}\log_2\frac{2e}{\varepsilon}\right\}
\quad(0<\varepsilon\leq2).}
$$
At zero the limit is zero. For small $\varepsilon$, the sharp exact answer behaves as $(\varepsilon/2)\log_2(2/\varepsilon)+\varepsilon/(2\ln2)+O(\varepsilon^2)$, displaying the logarithmic continuity of entropy near a <pure state>.
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