Let $f:A\rightarrow \prod_{\alpha \in J} X_{\alpha}$ be given by $f(a)=(f_{\alpha}(a))_{\alpha \in J}.$ Where $f_{\alpha}:A\rightarrow X_{\alpha}$ for each $\alpha.$ Then $f$ is continuous iff $f_{\alpha}$ is continuous for each $\alpha$ provided the product space $\prod_{\alpha \in J} X_{\alpha}$ is given product topology.
My attempt: ($\Rightarrow$)
Approach(1): let $x\in A$. Let $V_\alpha \in \mathcal{N}_{f_\alpha (x)}, \forall \alpha \in J$. Clearly $\prod_{\alpha} V_\alpha \in \mathcal{B}_p \subseteq \mathcal{T}_{p}$(product topology) and $f(x)= (f_{\alpha}(x)) \in \prod_{\alpha} V_\alpha$. So $\prod_{\alpha} V_\alpha \in \mathcal{N}_{f(x)}$. Since $f$ is continuous, $\exists Q\in \mathcal{N}_x$ such that $f(Q)\subseteq \prod_{\alpha} V_\alpha$. So $f(Q)=\{ f(z)\in \prod_{\alpha} X_\alpha |z\in Q\} \subseteq \prod_{\alpha} V_\alpha$. Thus $f(z)=(f_\alpha (z))\in \prod_{\alpha} V_\alpha, \forall z\in Q$. Which implies $f_\alpha (z)\in V_\alpha , \forall z\in Q$. Hence $f_\alpha (Q)\subseteq V_\alpha ,\forall \alpha \in J$. Thus $f_\alpha$ is continuous $\forall \alpha \in J$. Is this proof correct?
Approach(2): $\pi_{\beta}^{-1}(U_{\beta})= \prod_{\alpha} U_\alpha$, where $U_\alpha = X_\alpha ,\forall \alpha \in J\setminus \{ \beta \}$. $\pi_{\beta}^{-1}(U_{\beta})\in \mathcal{T}_p$, if $U_\beta \in \mathcal{T}_{X_\beta}$. Hence $\pi_{\beta}$ is continuous for all $\beta \in J$. Claim: $f_\beta =\pi_\beta \circ f$. Proof: $\pi_\beta \circ f (x)=\pi_{\beta}(f(x))=\pi_\beta ((f_\alpha (x))_{\alpha \in J})=f_\beta (x)$. $f_\alpha$ is composite of continuous maps, hence $f_\alpha$ is continuous forall $\alpha \in J$.
Approach(3): It is easy to check $f^{-1}(\prod V_{\alpha})=\bigcap_{\alpha \in J} f_{\alpha}^{-1}(V_{\alpha})$. Let $V_{\beta}\in \mathcal{T}_{X_\beta}$ and $\prod V_{\alpha} \in \mathcal{B}_p, V_\alpha =X_\alpha, \forall \alpha \in J\setminus \{ \beta \}$. So $f^{-1}(\prod V_{\alpha}) =f_{\beta}^{-1}(V_\beta)\cap A= f_{\beta}^{-1}(V_\beta)\in \mathcal{T}_{A}$. Hence $f_\beta $ is continuous for all $\beta \in J$. Is this proof correct?
($\Leftarrow $) Approach(1): Similar proof Theorem 18.4 of Munkres’ Topology don’t work. Precisely, if $Q_\alpha \in \mathcal{N}_x ,\forall \alpha \in J$, then it is not necessary that $\bigcap_{\alpha \in J}Q_\alpha$ is open in $A$. Hence this approach don’t work!
Approach(2): $f^{-1}(\prod V_{\alpha})=\bigcap_{\alpha \in J} f_{\alpha}^{-1}(V_{\alpha})= A\bigcap (\bigcap_{\alpha \in F} f_{\alpha}^{-1}(V_{\alpha}))$, where $F$ is a finite subset of $J$. Since $f_\alpha$ is continuous, $\bigcap_{\alpha \in F} f_{\alpha}^{-1}(V_{\alpha})\in \mathcal{T}_A$. Thus $f^{-1}(\prod V_{\alpha}) \in \mathcal{T}_A$.
Approach(3): $f^{-1}(\pi_{\beta}^{-1}(U_\beta))=(\pi_\beta \circ f)^{-1}(U_\beta)=f_{\beta}^{-1}(U_\beta)\in \mathcal{T}_{A}$.