Let $\{(X_\alpha,\mathscr T_{\alpha}):\alpha \in \Lambda\}$ be an indexed family of Hausdorff spaces such that each $X_\alpha$ has at least two points, and let $X=\Pi_{\alpha \in \Lambda} X_{\alpha}.$ Then $(X, \mathscr T)$ is separable. Then, $|\Lambda|\leq |\mathbb R|$ and $(X_\alpha,\mathscr T_\alpha)$ is seperable for each $\alpha \in \Lambda$.
$(X, \mathscr T)$ is separable then I could able to prove that $(X_\alpha,\mathscr T_\alpha)$ is seperable for each $\alpha \in \Lambda$.
How do I prove that defined $f$ is injective? Claim:-$|\Lambda|\leq |\mathbb R|$
For each $\alpha \in \Lambda, \{(X_\alpha,\mathscr T_{\alpha})\}$ be an indexed family of Hausdorff spaces such that each $X_\alpha$ has atleast two points. For each $\alpha \in \Lambda$, let $U_\alpha$ and $V_\alpha$ be disjoint members of $\mathscr T_\alpha.$ Let $A$ be the countable dense subset of $(X, \mathscr T)$. For each $\alpha \in \Lambda,$ let $A_\alpha=A\cap \pi^{-1}_{\alpha}(U_\alpha)$. Since $\overline A=X \implies A_\alpha \neq \emptyset, \forall \alpha \in \Lambda.$ Define $f:\Lambda \to \mathscr P(A)$ as $f(\alpha)=A_\alpha$. How do I prove that $f$ is injective?
My attempt:-$\alpha \neq \beta$. How do I prove that $A_\alpha \neq A_\beta$?
Let $x\in A_\alpha=A \cap \pi^{-1}_{\alpha}(U_\alpha)\implies x\in A$ and $f(x)\in U_\alpha\implies f(x) \notin V_{\alpha}.$ How do I proceed further?
[$\pi_{\alpha}: X\to X_\alpha$ defined as $\pi((x)_{\alpha \in \Lambda}))=x_\alpha$]
If $\alpha \neq \beta$, then $\pi_\alpha^{-1}[U_\alpha]\cap \pi_\beta^{-1}[V_\beta]$ is open and non-empty in $X$ and so intersects $A$ (which is dense), say: $$\exists p \in X: p \in A \cap \pi_\alpha^{-1}[U_\alpha]\cap \pi_\beta^{-1}[V_\beta]$$
But this $p \in f(\alpha)$ as $$A \cap \pi_\alpha^{-1}[U_\alpha]\cap \pi_\beta^{-1}[V_\beta] \subseteq \pi_\alpha^{-1}[U_\alpha] \cap A = A_\alpha = f(\alpha)$$ but $p \notin f(\beta)$ because $\pi_\beta^{-1}[V_\beta] \cap \pi_\beta^{-1}[U_\beta] = \emptyset$, so that $$ A \cap \pi_\alpha^{-1}[U_\alpha]\cap \pi_\beta^{-1}[V_\beta] \cap f(\beta) = \emptyset$$
This shows that $f(\alpha) \neq f(\beta)$ as sets, we have a member of one that's not in the other.
Considering $\pi_\alpha^{-1}[V_\alpha]\cap \pi_\beta^{-1}[U_\beta]$ we can find $p' \in f(\beta)\setminus f(\alpha)$ as well, just for fun.
This injection implies that $|\Lambda| \le |\mathcal{P}(A)|$ and if $A$ is countable (which we can choose to be the case if we assume $X$ is separable), we get $|\Lambda| \le 2^{\aleph_0} = \mathfrak{c} =|\mathbb{R}|$