Now, we must prove that for a collection $A_{\alpha}$ of subsets of $X_{\alpha}$, where each $A_{\alpha}$ is a subspace of $X_{\alpha}$. We need to prove that
$\prod_{\alpha \ \in \ J} \ A_{\alpha}$ is a subspace of $\prod_{\alpha \ \in \ J} \ X_{\alpha}$. Let us first define the subspace topology on each of the sets $A_{\alpha}$.
We have:
$$T_{A_{\alpha}} \ = \ \{ A_{\alpha} \ \cap \ U_{\alpha} \ | \ U_{\alpha} \ \in \ T_{\alpha}\}$$
Where $T_{\alpha}$ is the topology on $X_{\alpha}$. This means that all sets of the form $A_{\alpha} \ \cap \ U_{\alpha}$ will be open in each $A_{\alpha}$. Ok, now we need to
remember this in the next part of the proof. For the box topology, we know that a basis element is of the form:
$$B \ = \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}$$
Where each $U_{\alpha}$ is open in each of their respective topological spaces, $X_{\alpha}$, and $B$ is a basis element for $\prod_{\alpha \ \in \ J} \ X_{\alpha}$.
Now, if we want to form a basis for the topology on the set $\prod_{\alpha \ \in \ J} \ A_{\alpha}$ (the set we are trying to show has the subspace topology), then the basis elements will
be of the form that we saw above, but will be products of $V_{\alpha}$, with each $V_{\alpha}$ open in $A_{\alpha}$. We hav elaready shown that in general:
$$V_{\alpha} \ = \ A_{\alpha} \ \cap \ U_{\alpha}$$
With $U_{\alpha}$ open in $X_{\alpha}$. We then have as a basis element for $\prod_{\alpha \ \in \ J} \ A_{\alpha}$:
$$B \ = \ \displaystyle\prod_{\alpha \ \in \ J} \ V_{\alpha} \ = \ B \ = \ \displaystyle\prod_{\alpha \ \in \ J} \ (A_{\alpha} \ \cap \ U_{\alpha}) \ = \
\displaystyle\prod_{\alpha \ \in \ J} \ A_{\alpha} \ \cap \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}$$
And a union of basis elements is the general form of an open set:
$$U \ = \ \displaystyle\bigcup_{i} \ B^{i} \ = \ \displaystyle\bigcup_{i} \displaystyle\prod_{\alpha \ \in \ J} \ A_{\alpha} \ \cap \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}^{i} \ = \
\displaystyle\prod_{\alpha \ \in \ J} \ A_{\alpha} \ \cap \ \displaystyle\bigcup_{i} \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}^{i}$$
But this is the general form of an open set within the subspace topology for $\prod_{\alpha \ \in \ J} \ A_{\alpha}$ in $\prod_{\alpha \ \in \ J} \ X_{\alpha}$, and we have just proved that all
sets of this form are open in $\prod_{\alpha \ \in \ J} \ A_{\alpha}$, provided each $A_{\alpha}$ is a subspace of $X_{\alpha}$.
A similar argument applies for the product topology. We simply must consider the fact that a basis element in $\prod_{\alpha \ \in \ J} \ X_{\alpha}$ is given as:
$$B \ = \ \displaystyle\prod_{\alpha \ \in \ J} \ U_{\alpha}$$
Where, except for a finite number of indices, $U_{\alpha} \ = \ X_{\alpha}$. All we have to do is remember that if $A_{\alpha}$ has the subspace topology, then $A_{\alpha} \ \cap \ X_{\alpha}$
is open in $A_{\alpha}$, as $X_{\alpha}$ is open in $X_{\alpha}$. This means, that if we stipulate that when we have surpassed our finite indices, we have $V_{\alpha} \ = \ A_{\alpha} \ \cap \ X_{\alpha} \ = \ A_{\alpha}$,
then once we arrive at $U \ = \ \prod_{\alpha \ \in \ J} \ A_{\alpha} \ \cap \ \bigcup_{i} \ \prod_{\alpha \ \in \ J} \ U_{\alpha}^{i}$, each element of the form $\prod_{\alpha \ \in \ J} \ U_{\alpha}^{i}$ will be part of
the basis for $\prod_{\alpha \ \in \ J} \ X_{\alpha}$ in the product topology.