Question: Is this T or F? and most importantly, why?

I'll be using any answers for a basis or completely my other questions, since my understanding is still a little poor.

Answer: 
I had the same problem as you, and I'm writing this answer primarily for my own understanding. We work over the adele ring and idele group for a number field $K$, rather than just over $\mathbb Q$, but the idea is the same.

We aim to study the adelic topology on the ideles and understand why it is not generated by the same basis of open sets. 

Recall that, as abstract groups, $\mathbb I\_K\cong\mathbb A\_K^\times$. Now

$$
\mathbb A\_K^\times=\{\left(x\_v\right)\_{v\in M\_K}\colon x\_v\ne0\forall v,\left|x\_v\right|\_v=1\textrm{ for almost all }v\}
$$

A basic open set in the topology in the adeles is given by 
$$
\prod\_v U\_v
$$
where $U\_v$ is open in $K\_v$ for each $v$ and almost all $U\_v$ are equal to $\mathcal{O}\_{K\_v}$. 

Intersecting this with $\mathbb A\_K^\times$ we get the set
$$
\left\{\left(x\_v\right)\_{v\in M\_K}\colon x\_v\ne0\textrm{ for all }v,\;\left|x\_v\right|\_v=1\textrm{ for almost all }v,\;x\_v\in U\_v\textrm{ for all }v\right\}
$$
Now any such set is the union of sets of the form
$$
\prod\_v W\_v
$$
where $W\_v\subset K\_v^\times$ is open and $W\_v=\mathcal{O}\_{K\_v}^\times$ for almost all $v$. However, these sets aren't open in the subspace topology! The reason is actually fairly subtle: although any element $(x\_v)$ of our open set above satisfies $\left|x\_v\right|\_v=1$ for almost all $v$, these elements of absolute value $1$ can occur anywhere in the sequence $(x\_v)$ - i.e., the set of places $v$ such that $\left|x\_v\right|\_v\ne1$ can depend on the element $\left(x\_v\right)\_v$. The sets of the form $\prod\_v W\_v$ with $W\_v=\mathcal{O}\_{K\_v}^\times$ for almost all $v$ are only allowed to have elements $x\_v$ with norm not equal to $1$ in some *fixed* finite set of places $v$ independent of the individual elements.
Asking questions helps to clarify things... If I'm right, the mistake is that
$$(\Omega\_\infty-0)\times\dots\times(\Omega\_{p\_n}-0)\times(\mathbb{Z}\_{p\_{n+1}}-0)\times\dots\times(\mathbb{Z}\_{p\_{n+r}}-0)\times\mathbb{Z}\_{p\_{n+r+1}}^\times\times\mathbb{Z\_{p\_{n+r+2}}}^\times\times\dots=\mathbb{A}^\times\cap(\Omega\_\infty\times\dots\times\Omega\_{p\_n}\times\mathbb{Z}\_{p\_{n+1}}\times\dots\times\mathbb{Z}\_{p\_{n+r}}\times\mathbb{Z}\_{p\_{n+r+1}}^\times\times\dots),$$
is not open in the second topology since we end up with infinitely many $\mathbb{Z}\_{...}^\times$