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. $s \le 1$,
$$
\left|\sum_{\lvert k\rvert \le |x-y|^{-1}}
\hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le
\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert \lvert x-y\rvert \\ =
\lvert x-y\rvert \sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s} \\ \le
\lvert x-y\rvert \,
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2}
\left(\sum_{\lvert k\rvert \le |x-y|^{-1}}
\lvert k\rvert^{2s}\lvert \hat u_k\rvert^2 \right)^{1/2}\\
\lesssim \lvert x-y\rvert \, \|u\|_{H^s}
\left(\frac{2}{\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\
=2^{1/2}\|u\|_{H^s}\lvert x-y\rvert^{s-1/2}
$$
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 $0<s<1$
$$
\sum_{k=1}^n \frac{1}{k^s}\le \frac{(2-s)n^{1-s}}{s-1}.
$$
I think the last desired inequality cannot be right, but a direct counterexample escaped me, so here is an argument with a bit of a detour.
Let me maybe first frame this with an abstract result.
Lemma. If $A \colon X \supseteq D \to Y$ is a closed operator between Banach spaces with domain $D$, then $$\|Ax\|_Y \lesssim \|x\|_X \quad \text{for all}~x\in D$$ if and only if $D$ is a closed subspace of $X$. In particular, if $A$ is in fact densely defined, then the foregoing inequality is equivalent to $D = X$.
The "if" direction follows with the closed graph theorem. For the "only if" part, observe that if $D \supseteq (x_k) \to x$ in $X$, then by the assumed inequality, $(Ax_k)$ is a Cauchy sequence in $Y$ and thus convergent, hence the closedness of $A$ implies that $x \in D$.
That being said, consider $\gamma_t$ as an unbounded operator $D \subseteq H(\text{curl},\Omega) \to L^2(\partial\Omega)$ with the domain $$D = \Bigl\{ v \in H(\text{curl},\Omega) \colon \gamma_t(v) \in L^2(\partial\Omega)\Bigr\}.$$
Using the Green type formula, you can show that this is in fact a closed operator. Moreover, since, say, continuous differentiable functions on the closure of $\Omega$ are included in $D$, it is also densely defined. Hence, with the Lemma, it follows that the desired inequality $$\|\gamma_t(v)\|_{L^2(\partial\Omega)} \lesssim \|v\|_{H(\text{curl},\Omega)} \qquad \text{for all}~v \in D$$
is equivalent to $D = H(\text{curl},\Omega)$ which is, I suppose, certainly not correct.
Best Answer
This is not true in the case of $H^{-1}(\Omega) = H_0^1(\Omega)^*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded:
I would expect that the same arguments can be adopted to your situation.