Product of $T_1$ spaces is $T_1$

Question

I am trying to prove that the product of $T_1$ spaces is also $T_1$. Here is a proof, is it correct?

$\{ X_i \}_{i \in I}$ are T₁ $\Rightarrow$ $\prod_{i \in I} X_i$ is T₁

Proof: Let $\bar{x} = ( x_i ) \neq \bar{y} = (y_i) \in \prod_{i \in I} X_i$

There is an $i_0 \in I$ such that $x_{i_0} \neq y_{i_0} \in X_{i_0}$

$X_{i_0}$ is T₁, so there exists open sets $U_0,V_0 \subseteq X_{i_0}$ such that $x_{i_0} \in U_0 \setminus V_0$, $y_{i_0} \in V_0 \setminus U_0$.

Define $U,V \subseteq \prod_{i \in I} X_i$: $$U_j = \begin{cases} X_j, &j \neq i_0 \\ U_0, &j = i_0 \end{cases} \qquad V_j = \begin{cases} X_j, &j \neq i_0 \\ V_0, &j = i_0\end{cases}$$

$U,V$ are open in $\prod_{i \in I}X_i$ by definition, and $\bar{x} \in U \setminus V$, $\bar{y} \in V \setminus U$.

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Answer 1

Generally, for any family of spaces $(X_i)_{i\in I}$ and any family of subspaces $A_i\subset X_i$, $i\in I$ one has $$\overline{\prod_i A_i}=\prod_i \overline{A_i}$$ in particular, the product of any family of $T_1$-spaces is $T_1$, as all one point sets will be closed.