I'd like to add that if we assume $X$ to be a normed vector space (over $\mathbb R$) then we have $T_{norm} = T_{weak}$ if and only if $X$ is finite dimensional. To see why this is the case:
$\Longrightarrow$ Assume $X$ is infinite dimensional. To show that then $T_{norm} \neq T_{weak}$ it's enough to find a set that is closed in one of the two topologies but not in the other. Note that $S := \{x \in X \mid \|x\| = 1 \}$ is closed in $T_{norm}$. But it's not closed in $T_{weak}$ since $0$ is in the weak closure of $S$: Let $U$ be any neighbourhood of $0$ in $T_{weak}$. Then there exists an open set $O$ such that $0 \in O \subset U$. We know that $\bigcup_{\varepsilon>0, r_0 \in \mathbb R} \{ \bigcap_{i=1}^n f_i^{-1} (B(r_0, \varepsilon)) \mid n \in \mathbb N \}$ forms a neighbourhood basis of $T_{weak}$. Hence there exist $f_1, \dots f_n \in X^\ast, \varepsilon > 0, r \in \mathbb R$ such that $0 \in O = \bigcap_{i=1}^n f_i^{-1} (B(r, \varepsilon))$.
Now we define a map $\varphi : X \to \mathbb R^n$, $x \mapsto (f_1(x), \dots, f_n(x))$. This map is linear. Hence $\operatorname{dim}{X} = \operatorname{dim}{\ker \varphi} + \operatorname{dim}{\operatorname{im}{\varphi}}$. We know that its image has dimension at most $n$ so since $X$ has infinite dimension we know that its kernel has to have infinite dimension. In particular, we can find an $x$ in $X$ such that $x \neq 0$ and $f_i(x) = 0$ for all $i \in \{1, \dots n\}$, i.e., $\varphi (x) = 0$. Since $\varphi$ is linear we also have $\varphi (\lambda x) = 0$ for all $\lambda \in \mathbb R$, in particular, for $\lambda = \frac{1}{\|x\|}$. Hence we have found a point $\frac{x}{\|x\|}$ that is in $S$ and also in the neighbourhood $U$ of $0$. Since the neighbourhood $U$ was arbitrary we get that $0$ is in the weak closure of $S$.
$\Longleftarrow$ Let $X$ be finite dimensional. Since by definition we always have $T_{weak} \subset T_{norm}$ it's enough to show that every open ball $B_{\|\cdot\|}(x_0, \varepsilon)$ is weakly open. Since $X$ is finite dimensional we can write every $x$ in $X$ as $x = \sum_{i=1}^n x_i e_i$ where $e_i$ is the basis of $X$. Define $f_i : X \to \mathbb R$ as $f_i : x \mapsto x_i$. Then $f_i$ are in $X^\ast$. Also since $X$ is finite dimensional, all norms on $X$ are equivalent and hence it's enough to show that $B_{\|\cdot\|_\infty}(x_0, \varepsilon)$ is weakly open. But that is clear since
$ B_{\|\cdot\|_\infty}(x_0 , \varepsilon ) = \{x \in X \mid \max_i |x_i-x_{0i}|< \varepsilon \} = \{x \in X \mid \max_i |f_i(x-x_0)| < \varepsilon \} = \bigcap_{i=1}^n f_i^{-1} (B(x_{0i}, \varepsilon))$