Quote from the OP: Keeping this up since I put a lot of writing into it and it got starred. But it has an error which I will indicate in the post.
Define $C_n(A) = \{a \in A : \forall d \in \Bbb{N}$ one of $a + d, a +2d, \dots a + (n-1)d$ is not in $A\}$
For $n \leq m$, we have $C_n(A) \subset C_m(A)$.
Proof. Let $x$ be in the LHS. Then for all $d$, one of $a + d, a + 2d, a + 3d, \dots$ is missing from $A$. This makes the same true for the such lists in $C_m(A)$ since if one is missing in the shorter list then it is certainly missing in the longer list containing the shorter list. QED
Thus $C_n(A) \cap C_m(A) = C_{\min(m,n)}(A)$. Now use $\{C_n(A) : n \in \Bbb{N}\}$ to define a basis for a topology in $A$, namely unions of subsets of the form $C_n(A)$. We need to show that $A = \cup_{n\geq 1} C_n(A)$. If not, then for some $a \in A$, for all $k \in \Bbb{N}$ there exists $d$ such that $a, a+d, a+2d, \dots, a+(k-1)d$ are all in $A$. Now if arithmetic sequence length were bounded then we'd certainly have that this is so, and $A = \cup_{n\geq 1} C_n(A)$. So let $A$ be as in a statement of Erdos' conjecture on arithmetic sequences.. That is, let $A$'s reciprocal sum diverge and assume (by way of contradiction) that arithmetic sequence length in $A$ is bounded.
Notice that $C_n(A)$ contains no arithmetic sequence $\geq n$. Now let $\{C_i(A)\}$ be an open covering of $A$ in the above topology. Now let the maximal sequence length in $A$ be $K$. Then if $A \neq \cup_{k=1}^{K+1} C_k(A)$, then in some $C_{K+q}(A), q \gt 1$, there is $a$ such that for all $d$ one of $a + d, a + 2d, \dots, a + (K+q - 1)d$ is not in $A$. But if the smallest such number for some $d$ is at or above $a + Kd$, then there is a sequence of length greater than $K$, contradiction. Thus for each $d$ a list element that's not there is also one that is not there in the previous lists and so $a$ must be in one of the previous open balls. Therefore, a set with bounded arithmetic sequences is compact in this topology.
Now consider the topology presented here in which no infinite subset is compact. More specifically define a topology on $A$ by defining the closed sets to be convergent subsets of $A$ in the sense that $X$ is convergent if $\sum_{x \in X} \frac{1}{x}$ converges. Call this topology $(A, \tau_2)$ and the first topology described in this post call that $(A, \tau_1)$, then the identity map $\rm{id} : (A, \tau_1) \to (A, \tau_2)$ is continuous:
ERROR: clearly as $\rm{id}^{-1}(\{$ a convergent (closed) set $\}) = A - $ a divergent set with bounded arithmetic sequences $= A - $ compact subset of $A = A - \cup_{i}C_i(A) = $ a closed set in $\tau_1$. I wasn't able to show that such a divergent subset of $A$ is also compact
To finalize the proof, a continuous image of a compact set is compact, yet there are no infinite compact sets in $(A, \tau_2)$, a contradiction. Thus if $\sum_{a \in A} \frac{1}{a}$ diverges, then $A$ contains unbounded arithmetic sequences. That completes the proof.