When $X$ is a Banach space, in my textbook, the weak topology is defined as below: For every point $x_0 \in X$, finite subset $F \subset X^\ast$, positive real number $\varepsilon > 0$, we define $U(x_0, F, \varepsilon) = \lbrace x \in X \mid \forall \varphi \in F,\ |\varphi(x - x_0)| < \varepsilon \rbrace$. Weak topology of $X$ is the topology where for each $x_0 \in X$, $\lbrace U(x_0, F, \varepsilon) \mid F \subset X^\ast,\ \varepsilon > 0 \rbrace$ is a fundamental system of neighborhoods of $x_0$.
I'm a beginner of functional analysis so I tried specifying the weak topology of $\mathbb{C}$ as an exercise. Below is my proof and I found that the weak topology of $\mathbb{C}$ is the usual topology of $\mathbb{C}$, induced by Euclidean metric. Is my proof correct? And if not, please tell me what to be modified.
(Proof) I mean a functional $f_z \colon \mathbb{C} \to \mathbb{C}$ as $f_z(w) = zw$ ($w \in \mathbb{C}$) for every $z \in \mathbb{C}$. Note that the correspondence $\mathbb{C} \ni z \mapsto f_z \in \mathbb{C}^\ast$ is bijective, so finite subset $F \subset \mathbb{C}^\ast$ can be written as $F = \lbrace f_{z_1}, \ldots, f_{z_n} \rbrace$ ($z_1, \ldots, z_n \in \mathbb{C}$). Then, I found that if $\displaystyle r = \max_{1 \le i \le n} |z_i|$ is non-zero, $$ U(z_0, F, \varepsilon) = \bigcap_{i = 1}^n \lbrace z \in \mathbb{C} \mid |z_i||z - z_0| < \varepsilon \rbrace = \left\lbrace z \in \mathbb{C} \mid |z - z_0| < \frac{\varepsilon}{r}\right\rbrace. $$ And, I easily know $U(z_0, F, \varepsilon) = \mathbb{C}$ if $F$ is empty or $r = 0$. Since $\varepsilon > 0$ is arbitrary, $U(z_0) = \lbrace U(z_0, F, \varepsilon) \rbrace_{F, \varepsilon}$ contains all open balls centered at $z_0$, and therefore $U(z_0)$ is indeed a fundamental system of neighborhoods of $z_0$ with respect to the usual topology. A topology satisfying the condition is unique, so two are identical.