Product topology coarser than weak topology Let $\Gamma$ be a nonempty set and define the real vector space
$$
c_{00}(\Gamma):=\{f: \Gamma\to \mathbf{R} \text{ finitely supported}\}.
$$
Now, let $V$ be a vector subspace of the algebraic dual $\mathbf{R}^\Gamma$ and suppose that
$$
(c_{00}(\Gamma), V)
$$
is a dual pair (that is, a pair of vector spaces equipped with a bilinear map $\langle \cdot, \cdot \rangle: c_{00}(\Gamma)\times V\to \mathbf{R}$ such that both families of sections separate the points). Here, the duality map is $\langle x,y\rangle:=\sum_\gamma x(\gamma)y(\gamma)$.

Question. Is it true that the product topology $\tau$ on $c_{00}(\Gamma)$ is contained in the weak topology $\sigma(c_{00}(\Gamma), V)$?

The answer is affirmative in all examples I have in mind.
Attempt. Let $U$ be a basic $\tau$-neighborhood of zero, so that there exist $n\ge 1$, $\gamma_1,\ldots,\gamma_n \in \Gamma$, and $\varepsilon>0$ such that
$$
U:=\left\{x \in c_{00}(\Gamma): |x(\gamma_i)|\le \varepsilon \text{ for all } i=1,\ldots,n\right\}.
$$
Then, we have to show that $U$ is a weak neighborhood, that is, there exist $v_1,\ldots,v_k \in V$ and $\varepsilon_1,\ldots,\varepsilon_k>0$ such that
$$
\left\{x \in c_{00}(\Gamma): \left|\sum_\gamma x(\gamma)v_j(\gamma)\right|\le \varepsilon_j \text{ for all } j=1,\ldots,k\right\}\subseteq U.
$$
How to do it?
If the answer is affirmative, it should be related to the fact that $c_{00}(\Gamma)$ separates the points of $V$.
Ps. As it follows from Corollary 5.108 in "Infinite dimensional analysis" (Aliprantis and Border, 2006), $V$ separates the points of $c_{00}(\Gamma)$ if and only if $V$ is weak$^\star$-dense in the algebraic dual of $c_{00}(\Gamma)$.
Added (failed) attempt.
Denote by $\delta_{\gamma}$ the $\{0,1\}$-valued function $\Gamma\to \mathbf{R}$ such that $\delta_\gamma(z)=1$ iff $z=\gamma$.
Now, fix $U$ as above and a nonzero finitely supported sequence $x \in c_{00}(\Gamma)$ with
$$
\mathrm{supp}(x)\subseteq \{\gamma_1,\ldots,\gamma_n\}.
$$
Since $V$ is weak$^\star$-dense in the algebraic dual of $c_{00}(\Gamma)$, for each $i=1,\ldots,n$ there exist $v_i \in V$ such that
$$
\forall j=1,\ldots,n,\quad \left|\delta_{\gamma_i}(\gamma_j)-v_i(\gamma_j)\right|<\color{red}{\min\left\{\frac{\varepsilon}{4\sum_\gamma |x(\gamma)|}, \frac{1}{2}\right\}.}
$$
To conclude, fix $k=n$, and suppose $x$ satisfies
$$
\forall j=1,\ldots,n, \quad \left|\sum_\gamma x(\gamma)v_j(\gamma)\right|\le \varepsilon_j:=\frac{\varepsilon}{4}.
$$
Fix $j=i$ in the above inequality.
Then
$$
\frac{\varepsilon}{4} \ge \left|x(\gamma_i)v_i(\gamma_i)\right|-\left|\sum_{\gamma\neq \gamma_i}x(\gamma)v_i(\gamma) \right|=
\left|x(\gamma_i)v_i(\gamma_i)\right|-\left|\sum_{j\neq i}x(\gamma_j)v_i(\gamma_j) \right|
$$
$$
> |x(\gamma_i)| \left(1-\frac{1}{2}\right)-\sum_{j\neq i}|x(\gamma_j)|\cdot \frac{\varepsilon}{4\sum_\gamma |x(\gamma)|}
\ge \frac{|x(\gamma_i)|}{2}-\frac{\varepsilon}{4}.
$$
This implies that $|x(\gamma_i)|<\varepsilon$. By the arbitrariness of $x$, we conclude that $x \in U$.
(The mistake here is that the red number should be defined independently of the sequence $x$.)
 A: Suppose that $\Gamma$ is infinite and fix some $\gamma_0 \in \Gamma$.
Set $V$ to be the linear span of $\{e_\gamma; \gamma \in \Gamma \setminus \{\gamma_0\}\} \cup \{1\}$, where $e_\gamma$ are the canonical vectors and $1$ is the constant function.
Then $V$ separates point of $c_{00}(\Gamma)$: let $x \in c_{00}(\Gamma)$ satisfy $\langle x,v \rangle = 0$ for all $v \in V$. Then we have $$0 = \langle x, e_\gamma \rangle\ = x(\gamma); \; \text{ for all }\; \gamma \in \Gamma \setminus \{\gamma_0\}$$
and thus, as $1 \in V$, we also have
$$0 = \langle x,1 \rangle = \sum_{\gamma} x(\gamma) = x(\gamma_0).$$
Thus $x = 0$ and indeed $V$ separates points.
The fact that $\color{red}{\tau \not\subseteq \sigma := \sigma(c_{00}(\Gamma),V)}$ follows as $e_{\gamma_0}$ does not lie in $V$, but I will show this directly. It is enough to find a net $(x^\alpha)$ which converges in $\sigma$ but not in $\tau$. We will actually find a sequence: fix a sequence of distinct indices $(\gamma_n)$ in $\Gamma \setminus \{\gamma_0\}$ and take
$$x^n = (-1)^n e_{\gamma_0} + \frac{(-1)^{n+1}}{n} \sum_{k=1}^n e_{\gamma_k}.$$
Then $(x^n)$ converges to zero in $\sigma$ (it converges at each coordinate except $\gamma_0$ and $\langle x^n,1 \rangle = 0$ for all $n$) but it does not convege in $\tau$ (as $x^n(\gamma_0) = (-1)^n$ which does not converge).
