What is the closure of $(0,1)$ in $\mathbb{R}_k$? BACKGROUND
Let
$$K := \left\{\frac{1}{n} \mid n \in \mathbb{Z}_{+}\right\} = \left\{\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots\right\},$$
where $\mathbb{Z}_{+}$ is the set of all positive integers.
Let
$$\mathscr{B}_k = \left\{(a,b) \subseteq \mathbb{R} \mid a, b \in \mathbb{R}, a < b\right\} \cup \left\{(c,d)-K \mid c, d \in \mathbb{N}, c < d\right\}.$$
Then $\mathscr{B}_k$ is a basis for a topology on $\mathbb{R}$, and this is called the $K$-topology on $\mathbb{R}$ and is denoted as $\mathbb{R}_k$.
Now, a subset $C$ of the topological space $X$ is said to be closed if $X - C$ is open.  The closure of a set $A$ is the smallest closed set containing A.
QUESTION

What is the closure of $(0,1)$ in $\mathbb{R}_k$?

ATTEMPT
This would have been easy if the $K$-topology $\mathbb{R}_k$ is coarser than the standard topology $\mathbb{R}_s$.  However, we know that $\mathbb{R}_s \subset \mathbb{R}_k$.
Hence, I am unsure if the closure $\overline{(0,1)}$ of $(0,1)$ in $\mathbb{R}_k$ is still $[0,1]$ as it is in $\mathbb{R}_s$.
Any help will be appreciated.
 A: As the standard topology $\mathcal{T}_s \subset \mathcal{T}_K$ we have the same inclusion for closed sets as well. So $[0,1]$ is closed in $\mathbb{R}_K$. (So the closure of $(0,1)$ in $\mathcal{T}_K$ is still a subset of $[0,1]$.)
So we only need to check that $0$ and $1$ are still limit points of $(0,1)$ in the larger topology $\mathcal{T}_K$. The only point that has a different set of neighbourhoods (compared to the standard topology) is $0$, as $(a,b) - K$ is open in it, for intervals with $0 \in (c,d)$. 
How $\mathcal{T}_K$ was constructed: We have a non-closed set $K$, and $\mathbb{R}_K$ is the reals with the smallest topology larger than the usual one ,such that $K$ is closed in the larger topology. Before $0$ was the only point in $\overline{A} - A$, so only the neighbourhoods at $0$ need to change: we cut out $K$ so that there are lots of open sets that contain $0$ but do not intersect $K$ any more, like $(-1,1)-K$. But all $(c,d) - K$ where $0 \in (c,d)$ still intersects $(0,1)$ (say in some $\frac{\sqrt{2}}{n} \in (0,1)-K \subseteq (0,1)$ ,so $0 \in \overline{(0,1)}$ and the same holds for neighbourhoods of $1$ ; they are just the old open intervals essentially, so $1 \in \overline{(0,1)}$ as well.
So the closure is, as before, $[0,1]$. Alternatively show that $\frac{\sqrt{2}}{n} \to 0$ and $1-\frac{\sqrt{2}}{n} \to 1$ in $\mathcal{T}_K$, so that $0,1 \in \overline{(0,1)}$ and the use, as before, that $[0,1]$ is already closed.
A: Yes you are correct.

The closure of a set $A$ is  the set which consists of all the elements of $A$ and all the limit points of $A$
And also $cl(A)=\{B|B \supset A,B$closed $\}$

Every element in $(0,1)$ is a limit point of $(0,1)$ also $\{0,1\}$ are limit points of $(0,1)$ in this topology because every open neighbourhoud of $0$ and $1$ must contain points of $(0,1)$ different from  $0$ and $1$ respectively.
Also every point $x<0$ and $y>1$ is not a limit point of $(0,1)$.You can prove this,for instance for a $y>1$ by finding an interval $(a,b)$ which does not intersect $(0,1)$.Do the same for $x<0$
Note that  $[0,1]^c= (- \infty ,0) \cup (1, +\infty)$ and $$(- \infty ,0)= \bigcup_{n=1}^{\infty}(-n,0)$$ $$(1,+ \infty)= \bigcup_{n=2}^{+\infty}(1,n)$$
So the complement of $[0,1]$ can be written as a union of elements of the given basis so it's open with respect to the topology generated by it.
thus $[0,1]$ is closed.
Now closure of $(0,1)$ in this topology is $[0,1]$
Indeed  $cl[0,1] \subseteq [0,1]$ because closure is the smalest set containing $(0,1)$ and  $[0,1]$ is closed and contains $(0,1)$
Also every limit point of $(0,1)$ is in its closure so $[0,1] \subseteq cl((0,1))$
Thus $[0,1]=cl((0,1))$
