I think here is one example. For each rational in $[0, 1]$ of the form $k/2^{n}$, let
$$S_{k, n} := \{k/2^n\} \times \cup_{-2^n \leq j \leq 2^n} \{j2^{-2n}\}$$ and define
$$S = \bigcup_{k \in \mathbb Z, n \in \mathbb Z_+} S_{k, n} \cup ([0, 1] \times \{0\}).$$
Identifying the latter set in the union can be shown that the set of $x \in [0, 1]$ with the liminf in question $0$ for all $v \in \mathbb R^2$ is the set $E$ defined as follows:
For $z$ \in R, denote by $L_k (z)$ the length of the string of $0$’s or $1$’s beginning at the $k$’th decimal place (after the decimal point) of the binary expansion of $z$. Writing $(v_1, v_2)$ for $v \in \mathbb R^2$, and $\mathcal Q$ for the set of points with rational angles in the circle $S^1$, define $E_q$ for $q \in \mathcal Q$ by
$$E_q := \{x \in [0, 1] \times \{0\}| \ \limsup_{k \to \infty} \min_{i = 1, 2} L_k ([x + q2^{-k}]_i - k = +\infty\}.$$
Finally, set $E := \cup_{q \in \mathcal Q} E_q$.
I claim without proof that this set $E$ has full measure in $[0, 1]$, whence the set of $x \in S$. with $T_S (x) = \mathbb R^2$ is of $\mathcal H^1$ measure $1$.
Finally we note that $S$ is closed and has $\mathcal H^1$ measure $1$, and so $S$ satisfies the requirements of your problem.