When does the ability to get the end of a path arbitrarily close to node imply that the node is reachable from the start of the path? Example of a Non-Finite Edge Set
Suppose that $n$ is a whole number and that $ES$ is a non-finite subset of
$ \begin{Bmatrix}
    \text{} \\
    \{v, w\}: v, w \in \mathbb{R}^{n} \\
    \text{} \\
\end{Bmatrix} $
The set $ES$ represents the edge set of an non-finite graph.
For example, maybe $n = 1$ and $ES = \bigg\{\left\{\frac{1}{1}, \frac{1}{2}\right\}, \left\{\frac{1}{2}, \frac{1}{4}\right\}, \left\{\frac{1}{4}, \frac{1}{8}\right\}, \left\{\frac{1}{8}, \frac{1}{16}\right\}, \cdots, \left\{\frac{1}{2^{99}}, \frac{1}{2^{100}}\right\}, \cdots, \left\{\frac{-1}{2^{99}}, \frac{-1}{2^{100}}\right\}, \cdots, \left\{\frac{-1}{1}, \frac{-1}{2}\right\} \bigg\}$.
Or more formally, $ES = \left\{\left\{2^{-k}, 2^{-(k+1)}\right\}: k \in \mathbb{Z}\right\}$
Question, is $\frac{+1}{2}$ connected to $\frac{-1}{2}$?
Definition of "connected to"
INFORMAL
Informally, a vertex $v$ in a non-finite graph is connected to a vertex $w$ if and only if, no matter how we split the vertices into two non-empty sets, if $v$ is in one set and $w$ is in the other set, then there exists at least one edge between the two sets.
FORMAL
Suppose that $VS \subseteq \mathbb{R}$ such that $\bigcup\limits_{E \in ES}^{\text{ }} E$.
If $v \in \mathbb{R}^{n}$ and if $w \in \mathbb{R}^{n}$, then we say that $v$ is connected to $w$ if and only if for every set $VS$ subset of $\mathbb{R}^{n}$, if $v \in VS$ and $w \not \in VS$ then there exists $v^{\prime}, w^{\prime} \in \mathbb{R}^{n}$ such that $v^{\prime}  \in VS$ and $w^{\prime} \not\in VS$ and $\{v^{\prime}, w^{\prime}\} \in ES$
Remark About Connectivity
Note that if a dot named $y$ is connected to from a dot named $x$, then that does NOT imply that there exists a finite-length path from dot $x$ to dot $y$.
More formally, there exist non-finite graphs $G$ such that node $y$ is connected to node $x$ in $G$, and there does not exist a finite-length path from node $x$ to dot $y$.
Definition of "finitely approachable from"

$  \forall ES \subseteq \bigl \{\{v, w\}: v, w \in \mathbb{R}^{n} \bigr\} \\
\forall a, z \in {\bigcup\limits_{E \in ES}^{\text{ }} E} \subseteq \mathbb{R}^{n}, \quad \quad  \\
    \quad z \text{ is finitely approachable from } a \quad \\  
    \quad     \quad \text{ if and only if } \quad  \\   
    \quad     \quad {\forall \epsilon \in \{r \in \mathbb{R}: r > 0\}} \\
    \quad     \quad     \quad     \exists z^{\prime} \in {\bigcup\limits_{E \in ES}^{\text{ }} E} \text{ such that: }      \\ 
    \quad     \quad    \quad    \quad |z^{\prime} - z| < \epsilon              \\ 
    \quad     \quad  \quad\quad   \quad \text{and }           \\ 
    \quad     \quad \quad \quad\exists \text{ a finite length path } P \subseteq ES \text{ such that } P \text{ begins at node } a \text{ and ends at node } z^{\prime}       
$

Re-Phrasing my Question
If there exists a path from you to a node arbitrarily close to node $z$, what characterizes the conditions under which node $z$ is within in your reach?
I am not sure, but maybe my question could be re-phrased as "What is a useful way to fill in the blank $\bigl(\underline{  \quad \quad \quad }\bigr)$ in the following statement?"

$\forall a, z \in \mathbb{R}^{n}, \quad \quad  \\
    \quad z \text{ is connected to } a \quad \\  
    \quad     \quad \text{ if and only if } \quad  \\
    \quad \text{ both of the following two conditions are satisfied: }  \\  
    \quad     \quad \bigl(\underline{\quad \quad \quad \quad \quad \quad}\bigr)  \\ 
    \quad     \quad    \quad    \text{and} \\
    \quad     \quad z \text{ is finitely approachable from } a     
$

A Note
For a finite graph, it is trivial to say that one node is reachable from another.
We simply have a path between the nodes such that the path has finite length.
However, if you have a big graph, having a non-finite number of nodes, then things get tricky.
I am trying to find a theorem which makes it easy to say that some node is reachable from another node.
 A: Your claim

there exist non-finite graphs $G$ such that node $y$ is connected to node $x$ in $G$, and there does not exist a finite-length path from node $x$ to node $y$

is false. If there is no finite-length path from node $x$ to node $y$, then let $S$ be the set of all nodes $z$ such that there is a finite-length path from node $x$ to node $z$. Then

*

*$x \in S$: take the path of length $0$ starting and ending at $x$.

*$y \notin S$: by assumption.

*There is no edge $\{v,w\}$ such that $v \in S$ and $w \notin S$. If $v \in S$ and $\{v,w\}$ is an edge, we can take the finite-length path from $x$ to $v$ and then take one more step to $w$, getting a finite-length path from $x$ to $w$.

Therefore $x$ is not connected to $y$.
Whether your graph is finite or infinite, the condition "there is a path from $x$ to $y$" is the condition you want to say that $x$ is connected to $y$.

Now, on to your actual question - there is no useful condition that turns "finitely approachable from" into "connected". We could say that $a$ is connected to $z$ if

*

*$z$ is finitely approachable from $a$, and

*there is a ball of some radius $r>0$ around $z$ such that $z$ has an edge to every point in that ball.

This is overkill, but nothing less will do. Ultimately, until you add such a condition, there is nothing special about getting close to a point: edges don't care about distance.
