If adding ONLY finitely many points to a (non-empty) subset of plane, $\mathbb{R}^2$ makes it closed, then is the closure all of $\mathbb{R}^2$? Question: Let $E \subset \mathbb{R}^2$ be non-empty open set such that $E$'s union with finitely many points becomes a closed set. Is it necessarily true that
$$
\text{closure}{\ (E)} = \mathbb{R}^2 \, \, ?
$$
Equivalent Question: Looking into its complement you can rephrase it as: Given $F \subset \mathbb{R}^2$ closed, such that $F$ minus a finite set is a non-empty open set, does it follow that $ F = \mathbb{R}^2$ ?
Answer and Update: As shown by different correct answers below, yes, the closure of the set has to be all of $\mathbb{R}^2$. Also the proofs (at least many of them) illustrate the possibility of generalizing to the countable case, i.e.
Lemma: If an open set has only countably many boundary points, then its closure is all of $\mathbb{R}^2$.
Proof: Proof 1 (@Danielwainfleet's comment) Let $E$ be an open set whose complement is not all of $\mathbb{R}^2$. Pick a point in the complement of its closure which is a non-empty open set and consider an open ball around it that is contained in the complement. From its center every ray picks up a boundary point of $E$, and no two rays pick the same point, as the boundary point is not that center itself. This is where we use openness of the complement is used. This tells us that $E$ has uncountably many boundary points -- for the uncountably many rays.
(Other proofs do this with a line segment and that too will produce uncountably many boundary points.)
Proof 2: (Lee Mosher) Their proof below works if you replace the finite $A$ with the "countable" $A$ -- it is true that $\mathbb{R}^2 \setminus A$ is still path connected if $A$ is a countable set.
 A: Let $U$ be the given open, nonempty set, and let $A$ be the finite set, which we may assume is disjoint from $U$. 
The set $U \cup A$ is closed, so $V = \mathbb R^2 - (U \cup A)$ is open. It follows that $\mathbb R^2 - A = U \cup V$ is open. 
If $V$ were nonempty, it would follow that $\mathbb R^2 - A$ is disconnected. 
But, in fact, $\mathbb R^2 - A$ is path connected, as one can easily show. 
Therefore $V = \emptyset$ and $U \cup A = \mathbb R^2$.
A: The answer is yes. I start by providin a proof for the case where it is sufficient to add only one point, and then describe how to extend it to the general case.
We proceed by contradiction. Let $x_0\in \mathbb R^2\setminus\operatorname{closure}(E)$. Since $\operatorname{closure}(E)$ is closed, $\mathbb R^2\setminus\operatorname{closure}(E)$ is open and thus there exists $r>0$ such that $B_r(x_0)\cap \operatorname{closure}(E) = \varnothing$.
Since $E$ is open, there exists two points $y_1,y_2\in E$ such that $x_0,y_1,y_2$ are not aligned. (Otherwise $E$ is a subset of a line, which cannot be open.) 
Consider for $\lambda>0$ the point $z_1(\lambda) = (1-\lambda)x_0+\lambda y_1$. Since $|x_0-z_1(\lambda)|=\lambda|x_0-y_1|$, we know that $z_1(\lambda)\notin \operatorname{closure}(E)$ for $\lambda<r/|x_0-y_1|$ and $z_1(1)=y_1\in E$. By continuity of $\lambda\mapsto z_1(\lambda)$ there must exists $\lambda_1$ such that $z_1(\lambda_1)\in \operatorname{closure}(E)\setminus E$. The same argument on $z_2(\lambda)=(1-\lambda)x_0+\lambda y_2$ shows that there exists $\lambda_2>0$ such that $z_2(\lambda_2)\in \operatorname{closure}(E)\setminus E$.
Since $x_0,y_1,y_2$ are not aligned, we have $z_1(\lambda_1)\neq z_2(\lambda_2)$, and thus $\operatorname{closure}(E)\setminus E$ contains at least two points, which is a contradiction.
To derive the general argument, it suffices to apply the same idea to all points in a segment $[y_1,y_2]\subset E$ which is not contained in a line starting from $x_0$. This yields that if $\operatorname{closure}(E)\neq \mathbb R^2$ then $\operatorname{closure}(E)\setminus E$ contains infinitely many points.
A: I came up with an affirmative answer as follows. 
Lemma: Let $E \subset \mathbb{R}^2$ be open, and such that its union with a finite set is closed. Then $ \mathbb{R}^2 \backslash {E} $ is finite.
Proof: For each $ y \ $ define 
$$
E_y : = E \cap (\mathbb{R}^1 \times \{y\}) \, .
$$
Each $E_y$ is (basically) an open subset of $\mathbb{R}$.  If it is not all of $\mathbb{R}$ then it must have some boundary point. 
A boundary point of $E_y$ (as a subset of line), say $(x,y)$, is also a boundary point of $E$ as a subset of $\mathbb{R}^2$: Because every ball around $(x,y)$ picks a point $(x_1,y)$ from $E_y$ (hence from $E$), and a point $(x_2,y)$ from its complement (hence from complement of $E$).
Since by assumption $E$ has only finitely many boundary points, it follows that except for finitely many choices of $y$, 
$$
E_y = \mathbb{R} \, .
$$
And for those $y_1,y_2,\cdots,y_N$'s that do yield boundary points, since nearby lines are all contained wholly in $E$, every point in $\mathbb{R} \backslash E_{y_j}$ is a boundary point of $E$ -- (Careful! Not of $E_{y_j}$ necessarily.) To see this, if $(x,y_j) \notin E$, then the sequence $(x,y_j + 2^{-n})$ from $E$ (ignore first few terms) will converge to $(x,y_j)$, showing that $(x,y_j)$ is a boundary point of $E$.
Thus, even from these $\mathbb{R} \times \{y_j\}$ 's there are only finitely many not in $E$. 
Thus, we proved that $E$ is all of $\mathbb{R}^2$ except possibly for finitely many points. $\Box$
