Is a product of infinite subsets of $\Bbb C_p$ Zariski dense in $\Bbb C_p^2$? Consider the $p$-adic number field $\mathbb Q_p$. Let $\mathbb C_p:=\widehat{\bar{\mathbb Q}_p}$ with maximal ideal $\mathfrak{m}$.
Then any infinite subset $S \subset \mathfrak m$ is Zariski-dense in $\mathfrak{m}$.
Now consider a subset $T=T_1 \times T_2 \subset \mathbb C_p \times \mathbb C_p=\mathbb C_p^2$, where both $T_1$ and $T_2$ are infinite.

Is the infinite set $T$ a zariski-dense in $\mathbb C_p \times \mathbb C_p$ ?


We know that Zariski closure of a set is the intersection of the "zero-sets" of all polynomials vanishing on it. So Zariski closure $\bar T$ is equal to affine $2$-space $\mathbb C_p^2$ if and only if any polynomial vanishing on $T$ vanishes on $\mathbb C_p^2$. So we can think of $\bar T$ as the union of finitely many varieties. Then choose $V$ to be an irreducible component of $\bar T$ containing infinitely many points of $T$. Since $V$ is infinite, it is of positive dimension. So $\text{dim}(V)=1$ or $2$.
We have to show $\text{dim}(V) \neq 1$, in that case, $\text{dim}(V)=2$ and $V=\mathbb C_p^2=\bar T$.
Can you help me to conclude the proof ? Am I going into right direction ?
Thanks
 A: Here's an answer tying together all of the discussion here on this page.

Proof your set is dense
Let $K$ be an infinite field. Let $T_1,T_2\subset K$ be infinite subsets. Let $T=T_1\times T_2\subset K^2$. Then $T$ is dense in $K^2$ when $K$ is equipped with the Zariski topology. To prove this, we show that there is no non-zero polynomial in $K[x,y]$ which vanishes on $T$.
Suppose $f(x,y)$ vanishes on $T$. Without loss of generality, $f$ is of positive degree $d$ in $y$. Write $f(x,y)=\sum_{i=0}^d f_i(x)y^i$ with $d>0$ and $f_d(x)\neq 0$. Since $T_1$ is infinite and $f_d(x)$ has only finitely many roots, $f_d(x)$ cannot vanish on all of $K$ and therefore there is an $x_1\in T_1$ such that $f_d(x_1)\neq 0$. Therefore $f(x_1,y)$ is a polynomial in $y$ of degree $d$ which must vanish on the infinite set $\{x_1\}\times T_2$, contradiction.

Commentary on other solution methods
Note that one cannot appeal to an argument such as "the product of dense sets is dense in the product" because the Zariski topology on a product is not the product of the Zariski topologies (ref 1, ref 2, ref 3).
