Why does these following sets open or close? Could someone tell me if there is any mistake in my explanation of following statement?
(1) $(0,1)∪(2,4)$ is an open set in $\Bbb R$. 

Because for all points in the set we can find an open ball centered at the point.

(2) $[0,1]∪[2,4]$ is a closed set in $\Bbb R$ (do it in two diﬀerent ways).

Because 1.It includes all limit points.
          2.The complement is open.

(3) $(0,1]$ is neither open nor closed in $\Bbb R$. 

Because the left side does not contain the limit point 0 and also we cannot find an open ball centered at 1 that is contained completely in the set.

(4) Explain why $\Bbb R$ is both open and closed in $\Bbb R$. 

Because it contains all the limit points(by completeness) and for all points in $\Bbb R$ there is a neighbourhood that is completely contained in $\Bbb R$.

I fell not sure about the following 3. Could someone explain it explicitly just by using the following definition and theorem?
Deﬁnition. A set $A ⊂\Bbb R$ is open if for every $a ∈ A$ there exists $ε > 0$ such that the open interval $(a−ε,a + ε) ⊂ A$. 
Deﬁnition. A set $A ⊂\Bbb R$ is closed if its complement $\Bbb R\setminus A$ is open. 
Theorem. A set $A ⊂\Bbb R$ is closed if, and only if, for every sequence $(a_n)^∞_{n=1}$ in $A$ such that $a_n → a$, we have $a ∈ A$. That is, an equivalent deﬁnition for a closed set in $\Bbb R$ is that it contains all its limit points.
Deﬁnition. Let $(X,d)$ be a metric space. A set $A ⊂ X$ is open if for every $a ∈ A$ there exists $ε > 0$ such that the open ball $B(a,ε) ⊂ A$.
Deﬁnition. Let $(X,d)$ be a metric space. A set $A ⊂ X$ is closed if its complement $X\setminus A$ is open. 
Theorem. Let $(X,d)$ be a metric space. A set $A ⊂ X$ is closed if, and only if, for every sequence $(a_n)^∞_{n=1}$ in $A$ such that $a_n → a$, we have $a ∈ A$. That is, an equivalent deﬁnition for a closed set in $X$ is that it contains all its limit points.
(5) $(0,1)∪(2,4)$ is a closed set in the metric space $(0,1)∪(2,4)$. 
(6) $[0,1]∪[2,4]$ is an open set in the metric space $[0,1]∪[2,4]$. 
(7) $(0,1]$ is both open and closed in the metric space $(0,1]$.
Many thanks!
 A: All  cases you are unsure of (i.e., (5), (6), (7)) are a consequence of: 

If $X$ is a metric space then $X$ is both open and closed, and $\emptyset$ is both open and closed.

Note that $X$ is open because for every $a\in X$, the open ball $B(a,1)=\{\,y\in X\mid d(x,y)<1\,\}$ is a subset of $X$. And $\emptyset$ is vacuously open. Hence their complements $\emptyset$ and $X$ are closed.
A: Re: Equivalent def'n of "$A$ is closed" :
If $x$ is a limit point of $A,$  then for all $e>0$ we have $(-e+x,e+x)\cap A\ne \phi.$ So if $x\not \in A$ then $x\in \mathbb R$ \ $A$ and for all $e>0$ we have $(-e+x,e+x)\not \subset \mathbb R$ \ $A.$ So $\mathbb R$ \ $A$ is not open, so $A$ is not closed.
If $A$ contains all its limit points then for every $y\in \mathbb R$ \ $A$ there exists $e_y>0$ such that $(-e_y+y,e_y+y)\cap A=\phi.$ (Otherwise $y$ would be a limit point of $A,$ or a member of $A.$) So for every $y\in \mathbb R$ \  $A$ there exists $e_y>0$ such that $(-e_y+y,e_y+y)\subset \mathbb R$ \ $A.$ That is, $\mathbb R$ \ $A$ is open. So $A$ is closed.  
