= Solution
The small-argument <Taylor series> of the single-site response is
$$
\frac{2\sinh x}{1+2\cosh x}=\frac23x-\frac19x^3+O(x^5).
$$
Let $K=4\beta J$. For a small nonzero <order parameter> at zero field, the <self-consistency equation> becomes
$$
1=\frac{2K}{3}-\frac{K^3}{9}S^2+O(S^4).
$$
Since $K_c=3/2$ and $K=K_c/(1+t)$, where $t=T/T_c-1$, this gives
$$
0=-t-\frac38S^2+O(t^2,tS^2,S^4).
$$
Consequently
$$
\boxed{S_\pm=\pm\sqrt{\frac83}\,(-t)^{1/2}+O((-t)^{3/2}),\qquad t\to0^-.}
$$
The positive branch is the expression printed in the question, while its negative partner is required by <spin inversion symmetry>. The square-root behavior gives the <mean-field critical exponent> $1/2$ for the <spontaneous magnetization>.
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