I have produced an answer, not the most elegant one. 

Let $x\ne y$, then we have that
$$
\lvert u(x)-u(y)\rvert = \left|\sum_{k\in\mathbb Z}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|\le
\left|\sum_{\lvert k\rvert \le |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|+\left|
\sum_{\lvert k\rvert \ge |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|.
$$
We shall exploit the fact that 
$$
\big|\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big|\le
\min\big\{\lvert k\rvert\lvert x-y\rvert,2\big\}.
$$
For the first term we have two cases:

Case I. $s \le 1$,
$$
\left|\sum_{\lvert k\rvert \le |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le
\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert \lvert x-y\rvert =
\lvert x-y\rvert \sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s}  \\ \le 
\lvert x-y\rvert \,
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2}
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert k\rvert^{2s}\lvert \hat u_k\rvert^2 \right)^{1/2}
=\lvert x-y\rvert \, \|u\|_{H^s} 
\left(\frac{2}{\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\
=2^{1/2}\|u\|_{H^s}\lvert x-y\rvert^{s-1/2}
$$
Case II. $1<s<3/2$. We have
$$
\left|\sum_{\lvert k\rvert \le |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le
\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert \lvert x-y\rvert =
\lvert x-y\rvert \sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s}  \\ \le 
\lvert x-y\rvert \,
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2}
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert k\rvert^{2s}\lvert \hat u_k\rvert^2 \right)^{1/2} \\
=\lvert x-y\rvert \, \|u\|_{H^s} 
\left(\frac{4s}{(2s-1)\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\
=\left(\frac{4s}{2s-1}\right)^{1/2}\|u\|_{H^s}\lvert x-y\rvert^{s-1/2}
$$



For the second term we have
$$
\left|
\sum_{\lvert k\rvert \ge |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|\le 
2\sum_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u_k\rvert=2\sum_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert^s \lvert k\rvert^{-s}\\ \le 2\,
\left(\sum_{\lvert k\rvert \ge |x-y|^{-1}}\frac{1}{\lvert k\rvert^{2s}}\right)^{1/2}
\left(\sum_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u_k\rvert^2 \lvert k\rvert^{2s}
\right)^{1/2}\le 2\cdot\left(\frac{2\lvert x-y\rvert^{2s-1}}{2s-1}\right)^{1/2}\|u\|_{H^s} \\
=\frac{2^{3/2}}{(2s-1)^{1/2}}\cdot\lvert x-y\rvert^{s-1/2}\|u\|_{H^s}
$$
Altogether, for every $s\in(1/2,3/2)$, there exists a $c_s>0$, such that
$$
\lvert u(x)-u(y)\rvert\le c_s\lvert x-y\rvert^{s-1/2}\|u\|_{H^s},
$$
for all $u\in H^s(\mathbb T)$.


Note. We have used the following rather crude inequalities

a. For $s>0$,
$$
\sum_{k=1}^n k^s\le n^{s+1}.
$$

b. For $s>1$, 
$$
\sum_{k=n}^\infty \frac{1}{k^s}\le \frac{s}{(s-1)n^{s-1}}.
$$

c. For $0<s<1$
$$
\sum_{k=1}^n \frac{1}{k^s}\le \frac{(2-s)n^{1-s}}{s-1}.
$$