Show that $E = [0,1]$ is not open. Here is my proof. Not sure if I'm doing it correctly. We recall that $E$ is open provided every point of $E$ is an interior point. We then have $\forall p \in E$, there exists a neighborhood of $p$, $N$, such that $N \subset E$. Suppose that $E$ is open and consider the point $p=1$ in $E$. Since $E$ is open, then there exists a neighborhood, $N_r(1) = \{q : \ |q-1| < r\}$, such that $N_r(1) \subset E$. Here the topological space under consideration is $R$ with the usual real line topology.
Now, since $N_r(1) \subset E$ and $E$ is open, then each $p' \in N_r(1)$ is an interior point, that is, there exists a neighborhood, $N_{r'}(p') \subset E$. We will show that there exists a neighborhood, $N \subset N_r(1)$ for which $N \not\subset E.$ We consider three cases.
If $r \in (0,1)$, then we can take the neighborhood, $N_{r^2}(1+r^2)$. For example, is $r=\frac{1}{2}$, then
$$N_{\frac{1}{2}}(1)  =  \{q: \ |q-1| < \frac{1}{2}\} =\{q : \ q\in (\frac{1}{2}, \frac{3}{2})\},$$
$$N_{\frac{1}{4}}(5/4)  =  \{q: \ |q-\frac{5}{4}| < \frac{1}{4}\} =\{q : \ q\in (1, \frac{3}{2})\}.$$
Notice that $N_{\frac{1}{4}}(5/4) \subset N_{\frac{1}{2}}(1)$, but $N_{\frac{1}{4}}(5/4) \not\subset E$. On the other hand, if $r=1$, then we can take $N_{\frac{1}{4}}(\frac{3}{2})$, where
$$N_{\frac{1}{4}}(\frac{3}{2}) = \{q: \ |q-\frac{3}{2}| < \frac{1}{4}\} =\{q : \ q\in (\frac{5}{4}, \frac{7}{4})\}.$$
Clearly, $$N_{\frac{1}{4}}(\frac{3}{2}) \subset N_1(1) = \{q:  |q-1| < 1\} =\{q : \ q\in (0, 2)\},$$ but again $N_{\frac{1}{4}}(\frac{3}{2}) \not\subset E$. Finally, for $r > 1$, We can take $N_{\frac{1}{r^2}}(1+\frac{1}{r^2})$. For example, if $r=2$, then $N_2(1) =  \{q : |q-1|<2\} = \{q: q \in (-1,3)\}$ and $N_{\frac{1}{4}}(\frac{5}{4}) \subset N_2(1)$. Again, we have that $N_{\frac{1}{4}}(\frac{5}{4}) \not\subset E$. Does this argument shows that $E$ is not open?
 A: This is way too complicated. Let $N_r(1)$ be any basic nbhd of $1$. Then $1+\frac{r}2\in N_r(1)\setminus E$, so $N_r(1)\nsubseteq E$. Thus, $1$ is a point of $E$ that is not an interior point of $E$, so $E$ is not open.
A: Am I missing something? Just show that any open ball of positive radius with center $1$ or $0$ is not a subset of $[0,1]$
A: Assume $[0,1]$ is open. Then by definition there exists some $r > 0$ where we have that $|y - 1| < r$ implies $y\in [0,1]$. Choose $y = 1 + r/2$ then we have
$$|y - 1| = |r/2| < r$$
but $y\notin [0,1]$ so we have a contradiction.
A: One solution is to use the connectedness of $\mathbb{R}$. If $[0,1]$ is seen as a subset of the real line with the usual topology, $[0,1]$ is not open because $\mathbb{R}$ is connected and $\mathbb{R}\setminus[0,1]=(-\infty,0)\cup(1,\infty)$ is open (union of open sets is open).
The upshot:
If $[0,1]$ were open, then $\mathbb{R}$ would be the the disjoint union of two open sets, and so $\mathbb{R}$ would not be connected (contradiction to the well known fact that intervals, finite or infinite of the real line -with the usual topology- are connected.)

Another solution is to notice that $\{0,1\}$ are cluster points in  $[0,1]$ which are contained in $[0,1]$. Any neighborhood $(-\varepsilon,\varepsilon)$ of $0$ and $(1-\varepsilon,1+\varepsilon)$ fo $1$ containes elements of  $\mathbb{R}\setminus[0,1]$. That is, points $\{0,1\}$ are not interior points of $[0,1]$; thus $[0,1]$ is not open. (A set $U$ is open iff for any point $x\in U$ there is a basic open set $V$ such that $x\in V\subset U$).
A: By the definition of open set. You can not find neighborhood around $0$ and $1$ which contained in $E$. I.e., it is not possible to find neighborhood $N$ around $1$ such that $1\in N\subset E$.
