show that the interval of the form $[0,a)$ or $(a, 1]$ is open set in metric subspace $[0,1]$ but not open in $\mathbb R^1$ 
On the metric subspace $S = [0,1]$ of the Euclidean space $\mathbb R^1 $, every interval of the form $A = [0,a)$ or $(a, 1]$ where $0<a<1$ is open set in S. These sets are not open in $\mathbb R^1$

Here's what I attempted to show that $A$ is open in $S$. I have no idea how it is not open in $\mathbb R^1$.
Let $M = \mathbb R^1$, $x \in A  = [0,a)$ .
If $x = 0$, $ r \leq \min \{a, 1-a\}, \\\  B_S(0; r) = B_M(0,r)\cap[0,1] = (-r, r) \cap[0,1] = [0,r) \subseteq A $
If $x \neq 0, r \leq \min \{x,|a-x|, 1-x\},  \\B_S(x; r) = B_M(x,r)\cap[0,1] = (x-r, x+r) \cap[0,1] \subset (0,x) \text{ or } (x, 1) \subset A$
 A: The open sets in $[0,1]$ using the subspace topology are those of the form $U \cap [0,1]$ where $U$ is open in $\mathbb{R}$.
$[0,x)$ is open because, for example, it can be written as $(-1,x) \cap [0,1]$ and $(x,1]$ is open because it can be written as $(x,2) \cap [0,1]$.
$[0,x)$ is neither open  in $\mathbb{R}$ because every open set containing $0$ contains points  in $(-\infty,0)$, and is not closed because $x \notin [0,x)$, but $(x-\frac{1}{n}) \to x$, and $x-\frac{1}{n} \in [0,x)$ forall $n$ sufficiently large.
Similarly $(x,1]$ is not open. (Explicitly, the map $y \mapsto 1-(\frac{1-x}{x}) y$ is a homeomorphism of $[0,x)$ onto $(x,1]$.)
A: Hint: To show $(x,1]$ not open in $\mathbb R$ simply show that there is no open neighborhood of $1$ included in the half-closed interval.
A: Recall that open sets in $[0,1]$ can be characterized as all intersections of open sets in $\mathbb R$ with $[0,1]$. If you don't know this result, consult your textbook. Can you find an open set $U$ so that $U \cap [0,1]$ is $[0,x)$? What about $(x,1]$?
To show these half-open intervals are not open in $\mathbb R$, recall what is means to be open in $\mathbb R$. Find a point so that no neighborhood of the point is contained in the set. For example, if you have $[0,x)$, look at neighborhoods of $0$. 
(This answer assumes you are using the subspace topology on $[0,1]$. If you don't know what this means, ignore this parenthetical comment.)
