What you seem to be asking is this: If $C$ is a subset of $\mathbb{R}$, such that for any convergent sequence of elements of $C$, the limit is also an element of $C$, then must $C$ be closed?
The answer is Yes; topological spaces that satisfy this condition are called "sequential spaces," and $\mathbb{R}$ is a sequential space. Here's why:
Suppose that $x\in\mathbb{R}$, and that every neighborhood of $x$ intersects $C$. Then for each natural number $n$, let $x_n$ be a point in the intersection of $C$ with $(x-\frac 1n, x+\frac 1n)$. Then $x_n\to x$, so $x\in C$. Therefore $C$ must be closed.
This argument works whenever you replace $\mathbb{R}$ with a "first-countable" topological space (each point admitting a countable neighborhood basis, here given by $\{(x- \frac 1n, x+\frac 1n): n\in\mathbb{N}\}$). There are, however, sequential spaces which are not first-countable.