Every Interval of the form $[0,x)$ where $0I am confused on how to show that:
In the metric subspace $S=[0,1]$ of Eucledian Space $R^1$, every interval of the form $[0,x)$ where $0<x<1$ is an open set in $S$ by using just definitions. 
This question was posted already before but the answers attack for the proof is by using the theorem:
If $(S,d)$ is a metric subspace of $(M,d)$ and $X$ is a subset of $S$.Then $X$ is open in $S$ if and only if $X=A\cap S$ for some open set $A$ in $M$. Which is relatively easy than by using definition. 
Could someone help me to prove the problem by using just definitions? 
Here is my attempt: (Note:A  set $Y$ is open in $S$ if all its points are interior).
Let $Y=[0,x)$ and $S=[0,1]$. I must show that all the points in $Y$ are interior relative to $S$. This means that for all $y\in Y$ there is some open interval $B_s(y,r)$ that lies entirely in $Y$. A complication occurs by questioning what if $y=0$, I cant find an open interval $B_s(0,r)\subset Y$.  
[UPDATE: Got it now, thanks everyone.] 

The "theorem" you use is actually the definition of the induced topology on a subset of a topological space. Now for the actual exercise, let $0<x<1$ be given. Now $(-x,x)$ is open in $\mathbb{R}$, hence $(-x,x)\cap S=[0,x)$ is open in $S$.
