Perhaps, if someone could help me visualize $2^{52}$. Like how big is this number if we shuffle a single deck, should we expect this to be a new order or not? I know this is more of the soft part of the question.

Assuming that by $\chi_0$ you mean the trivial Dirichlet character then you are correct; however, for other characters we do know some things. For example for $\chi$ is a complex character (that is it takes values outside of $\mathbb{R}$, then there will not be any Siegel zeros. But if $\chi$ is real (say a quadratic character), then the existence of Siegel zeros is still open. But one could then talk about general $L$-series attached to other objects such as autommorphic forms

Perhaps, there should also be a more general comment here that for a general Dirichlet $L$ series $L(s,\chi)$ where $\chi$ is a Dirichlet character, the question of a real zero (known as a Siegel zero) is a difficult unresolved problem.

Yes, we do know that there are no nontrivial zeros on the real line. You might want to see arxiv.org/abs/2004.09765 which shows that there is no nontrivial zeros $s=\sigma+it$ off of the critical line for $\vert t\vert\leq 3\times 10^{12}$

I think my question is more about the topology whose closed sets are generated by zero sets of polynomials of degree $1$.

I don't see anything wrong with this proof it seems fine to me as long as the fixed point theorem doesn't somehow use this fact in its proof (which I don't believe it does), so I think you are fine

If you read the comment in the question you referenced they give an example of what the issue is, so perhaps you should read that discussion

Yes since if $f-g\in I$, then we have that $f-g=h$ for some $h\in I$, so then $f=h+g$, and as $g,h\in I$, we then have that $h+g=f\in I$ a contradiction

Perhaps, I might note that you can observe that the ideal $(2)$ factors as $(2)=(1+\sqrt{3})^2$, as you may observe that $2=(-1+\sqrt{3})(1+\sqrt{3})$.

Some will define a partition of a set $X$ to be a set $\{X_i\}_{i\in I}$ where $X_i\cap X_j=\emptyset$ for $i\neq j$ and $\cup_{i\in I}X_i=X$, so with this definition the empty set is allowed to be a part in the partition, so perhaps just take that as a part of the definition.