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 $0<s<1$
$$
\sum_{k=1}^n \frac{1}{k^s}\le \frac{(2-s)n^{1-s}}{s-1}.
$$
The answer is yes, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on Wikipedia.
The details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T_n)$ with a general subset $\mathcal{T} \subseteq \mathcal{L}(X)$.
Proof. By Baire's Theorem we can find an integer $m$ such that the set
\begin{align*}
B := \{x \in A: \, \|Tx\| \le m \text{ for all } T \in \mathcal{T}\}
\end{align*}
has non-empty interior within $A$. Thus, we can find a point $x_0 \in B$ and a real number $\varepsilon \in (0,1]$ such that each $x \in A$ which has distance at most $\varepsilon$ to $x_0$ is contained in $B$.
Now, let $y \in A$. The vector $z := x_0 + \frac{\varepsilon}{2}(y-x_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x_0$ is at most $\varepsilon$ since both $y$ and $x_0$ have norm at most $1$. Thus, $\|Tz\| \le m$ for all $T \in \mathcal{T}$. Since
\begin{align*}
y = \frac{2}{\varepsilon}(z - x_0) + x_0,
\end{align*}
we conclude that
\begin{align*}
\|Ty\| \le \frac{4m}{\varepsilon} + m
\end{align*}
for all $T \in \mathcal{T}$. This bound does not depend on $y$.
Best Answer
You can duplicate the usual proof of Hardy type inequalities to the discrete case.
Suppose $\{q_n\}$ is an eventually 0 sequence (you can weaken this to $\lim_{n\to \infty} n^{1/2} q_n = 0$). Then by telescoping you have (all sums are over $n\geq 0$)
$$ \sum (n+1) q_{n+1}^2 - n q_{n}^2 = 0 $$
Rewrite as
$$ \sum q_n^2 = - \sum (n+1) (q_{n+1} + q_n) (q_{n+1} - q_n) $$
Take absolute values and apply Cauchy-Schwarz on the RHS
$$ \sum q_n^2 \leq \left( \sum (n+1)^2 |q_{n+1} - q_n|^2 \right)^{1/2} \left( \sum (q_{n+1} + q_n)^2 \right)^{1/2} $$
the second factor can be bounded by $2 ( \sum q_n^2)^{1/2}$. Cancelling and you get
$$ \sum q_n^2 \leq C \sum (n+1)^2 |q_{n+1} - q_n|^2 $$
Scott Armstrong's comment is similar. As long as $\lim_{n\to\infty} q_n = 0$ you have
$$ q_n = \sum_{k \geq n} q_k - q_{k+1} $$
then
$$ \sup_{n\geq 0} |q_n| \leq \sum_{n\geq 0} |q_n - q_{n+1}| $$
If you want to look at "Sobolev type" inequalities: they are all essentially based on the fundamental theorem of (discrete) calculus applied in various ways.
Finally: note that you can also do a scaling argument.
Let $\lambda$ be a positive integer.
Let $q^{(\lambda)}_{n}$ be such that $$ q^{(\lambda)}_m = q_n \text{ if } m \in [\lambda n, \lambda (n+1))$$
This scaling preserves the $ \sum |q_{n+1} - q_n|^2$ semi-norm, but has $ \sum |q^{(\lambda)}_n|^2 = \lambda \sum |q_n|^2$, which immediately falsifies your proposed Poincaré inequality.
You see that the inclusion of weights in Hardy avoids this difficulty.