No, your argument is not correct. I’ll comment on it in detail.
Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point.
While it is true that every point of $[0,1]$ is a limit point of $[0,1]$, this is not a matter of definition: there is nothing in the definition of closed set that requires every point of a closed set to be a limit point of that set. Moreover, $[0,1]$ isn’t a subset of $\Bbb R^2$. I suspect that you’re thinking of the set $[0,1]\times\{0\}$, the set of points on the $x$-axis between $0$ and $1$ inclusive.
It is proved that all points in $R^2$ are limit points.
Limit points of what? They’re all limit points of $\Bbb R^2$, but that has nothing to do with $E$, which isn’t even a subset of $\Bbb R^2$. And if your $E$ was really supposed to be $[0,1]\times\{0\}$, then it’s not true that all points of $\Bbb R^2$ are limit points of $E$: if they were, they’d have to be in $E$, since $E$ is closed, and a closed set contains all of its limit points.
Therefore $E^c$ must contain all limit points.
Again, all limit points of what? And what does $\Bbb R^2$ have to do with $E^c$? $E^c=(\leftarrow,0)\cup(1,\to)$, a subset of $\Bbb R$, not of $\Bbb R^2$. For that matter, why are you bringing in $E^c$ at all? The question is whether every point of $E$ is a limit point of $E$, and $E^c$ isn’t going to help you to answer that question.
But $E^c$ by definition is open ($E^c=R^2\setminus E$).
See above: $E^c=\Bbb R\setminus E$, not $\Bbb R^2\setminus E$.
This is a contradiction because $E^c$ has 2 limit points,
No, every point of $(\leftarrow,0]\cup[1,\to)$ is a limit point of $E^c$; that’s far more than two limit points!
namely $E=[0,1]$.
No, the only points of $E$ that are limit points of $E^c$ are $0$ and $1$; the points of $E$ that are in $(0,1)$ are not limit points of $E^c$.
Therefore E contains points which are not limit points.
This doesn’t follow from anything that you’ve said, and in fact it contradicts your very first claim about $E$ up above. Moreover, it isn’t true, assuming that you’re talking about limit points of $E$: every point of $E$ is a limit point of $E$.
In order to answer the question, you must do one of two things: either prove that whenever $E$ is a closed subset of $\Bbb R^2$, then every point of $E$ is a limit point of $E$, or find an example of a closed subset $E$ of $\Bbb R^2$ containing a point that is not a limit point of that set $E$. You won’t be able to do the former, because it isn’t true. For the latter, just take any non-empty finite subset of $\Bbb R^2$, for instance $E=\{\langle 0,0\rangle\}$: $E$ is closed, and it has no limit points at all. In particular, $\langle 0,0\rangle$ is not a limit point of $E$, because it is not true that every open neighborhood of $\langle 0,0\rangle$ contains a point of $E$ different from $\langle 0,0\rangle$. If you want $E$ to be an infinite set, you can use $E=\big(\Bbb R\times\{0\}\big)\cup\{\langle 0,1\rangle\}$, for instance. This is a closed set, but $\langle 0,1\rangle$ is a point in $E$ that is not a limit point of $E$: if $\epsilon\le 1$, the open ball of radius $\epsilon$ centred at $\langle 0,1\rangle$ does not contain and point of $E$ besides the point $\langle 0,1\rangle$ itself. For a more extreme example, let $E=\{\langle m,n\rangle:m,n\in\Bbb Z\}$. You should try to prove that $E$ is an infinite closed set that has no limit points at all.
A closed set always contains all of its limit points. That is, suppose that $E$ is closed. If $p$ is a limit point of $E$, then $p\in E$. The point of this exercise is to show that the converse is not true: there can be a point $p\in E$ that is not a limit point of $E$.
Best Answer
(1).Your answers are correct and your reasoning has no flaws.
(2).The usual notation for an open ball, of radius $e$, centered at $x$, in a metric space $(X,d)$, is $B_d(x,e)$. When only one metric ($d$) is under consideration the subscript "$d$" is often omitted : $B(x,e)$. (Caution: In general we can have $B_d(x_1,e_1)=B_d(x_2,e_2)$ with $x_1\ne x_2$ or $e_1\ne e_2 $ or both.)
(3).Your def'n of relatively open (your last paragraph) is flawed: For metric space $(X,d)$ and $A\subset D\subset X,$ the set $A$ is relatively open in $D$ iff $$\forall x\in A \;\exists e>0\;(B_d(x,e)\cap D\subset A).$$
(4). A general approach: Let $T$ be a topology (the set of open sets) on a set $X.$ (Look up, if necessary, the general def'n of a topology). Closed sets are defined as the complements of open sets.
(5).In a metric space, the topology defined by (generated by ) the metric is: A set is open iff it is the union of a set of open balls. We can deduce that in a metric space a set is closed iff it contains all its metric limit points.
(6).Let $T$ be any topology on a set $X.$ Let $Y \subset X.$ The subspace topology on $Y,$ with respect to the topology $T,$ is $$\{t\cap Y: t\in T\}.$$ From this it is immediate that if $Z\subset Y$ then $Z$ is relatively open in $Y $ iff $Z=t\cap Y$ for some $t$ that is open in $X.$ (In particular if $Z\subset Y$ and $Z$ is open in $X$ then $Z$ is relatively open in $Y$.)
Deduce that for a metric space, this def'n of relatively open is equivalent to the def'n of relatively open in (3).
(7). For $S\subset X,$ the closure of $S$ in $X,$ sometimes denoted $Cl_X(S)$, is defined as the intersection of all closed subsets of $X$ that have $S$ as a subset. Equivalently, $X$ \ $Cl_X(S)$ is the union of all open subsets of $X$ that are disjoint from $S$.
And $S$ is closed in $X$ iff $S=Cl_X(S)$.
For $S\subset Y\subset X,$ the closure of $S$ in $Y$, denoted $Cl_Y(S)$ is the intersection of all subsets of $Y$ that are relatively closed in $Y$ and contain S as a subset. And if $S \subset Y$ then $S$ is closed in $Y$ iff $S=Cl_Y(S).$
An important, useful deduction is $$Cl_Y(S)=Y\cap Cl_X(S) \;\text {for all }\; S\subset Y.$$ In particular, if $Y\supset S=Cl_X(S)$ then $S$ is relatively closed in $Y$.
(8). Finally, with $X=\mathbb R$ and $Y=[0,2)$ we have:
(i)...$[0,1]$ is closed in $X$ so it is closed in $Y$. If $t$ is an open subset of $X$ and $t\supset [0,1]$ then $t\supset [1,1+r)$ for some $r\in (0,1).$ So $t\cap Y\supset [1,1+r)$ so $t\cap Y\ne [0,1].$ So $[0,1]$ is not open in $Y.$
Or we may observe that $Cl_Y(Y$ \ $[0,1])=Y\cap Cl_X((1,2))=Y\cap [1,2]=[1,2)\ne Y$ \ $[0,1]$, so therefore $Y$ \ $[0,1]$ is not closed in $Y$.
(ii)...The same argument that shows that $[0,1]$ is not open in $Y$ will also show that $(0,1]$ is not open in $Y.$ And $Cl_Y((0,1])=Y\cap Cl_X((0,1]=Y\cap [0,1]=[0,1]\ne (0,1]$ so $(0,1]$ is not closed in $Y.$
(iii)...$(-1,1)$ is open in $X$ so $(-1,1)\cap Y=[0,1)$ is open in $Y$. And $Cl_Y([0,1))=Y\cap Cl_X([0,1))=Y\cap [0,1]=[0,1]\ne [0,1)$ so $[0,1)$ is not closed in $Y.$
(iv). Since $[0,1)$ is open but not closed in $Y ,$ its complement in $Y,$ which is $[1,2),$ is closed but not open in $Y$.