Infinite intersection of an interval and probability of selecting a random point I am attempting to solve the following:

Let $A_n= (\frac{1}{2}-\frac{1}{2n}, \frac{1}{2}+\frac{1}{2n})$. Show that  $\bigcap\limits_{n=1}^\infty A_n=\{\frac{1}{2}\}$.  Then apply the continuity property of the probability function to show that the probability of selecting $\frac{1}{2}$ in a random selection from $(0,1)$ is zero.

My attempt: note that $(A_n)_n$ is a decreasing sequence of intervals hence $\bigcap\limits_{i=1}^n A_{i}= A_{n}$ for every $n$. So  $\bigcap\limits_{i=1}^\infty A_{i}= \lim\limits_{n\to\infty}\bigcap\limits_{i=1}^n A_{i}=\lim\limits_{n\to\infty}A_{n}=(\frac{1}{2}-0, \frac{1}{2}+0)=\{\frac12\}$.  Not sure if this is the right way to go about it.
I have no idea how to start the next part.  I know that if $(x_n)_n$ is a convergent sequence, then a function $f$ is continuous if $\lim\limits_{n\to\infty}f(x_{n})=f\left(\lim\limits_{n\to\infty}x_{n}\right)$, and that the probability function is continuous.
I guess I am confused on how we show that because the limit of an infinite intersection of intervals is a singleton set, $\{\frac12\}$, that the probability of selecting $\frac12$ from the interval $(0,1)$ is zero.
 A: Since $A_n\supset A_{n+1}$ for all $n$ and  $$A_n\stackrel{n\to\infty}\longrightarrow \left\{\frac12\right\},$$ by continuity from above it follows that 
$$ \mathbb P(A)=\mathbb P\left(\bigcap_{n=1}^\infty A_n \right) = \lim_{n\to\infty}\mathbb P(A_n)=\lim_{n\to\infty} \frac1n = 0.$$
A: Proving that no positive number is $<\dfrac 1 {2n}$ for every positive integer $n$ is a mathematical analysis problem that has been discussed on math.stackexchange.com in other questions.
If some number other than $1/2$ is a member of $\displaystyle \bigcap_{n=1}^\infty \left(\frac{1}{2}-\frac{1}{2n}, \frac{1}{2}+\frac{1}{2n}\right)$, then the distance between that number and $1/2$ is $<1/(2n)$ for every positive integer $n$.  But if it's a number other than $1/2$ then the distance is a positive number.
No matter what the probability distribution is, the probability assigned to $\{1/2\}$ must be $\le$ the probability assigned to $\left(\frac{1}{2}-\frac{1}{2n}, \frac{1}{2}+\frac{1}{2n}\right)$, no matter how big $n$ is.  "The continuity property" must be the cumulative distribution function is continuous.  The probability assigned to $\left(\frac{1}{2}-\frac{1}{2n}, \frac{1}{2}+\frac{1}{2n}\right)$ is $\le$ the probability assigned to $\left(\frac{1}{2}-\frac{1}{2n}, \frac{1}{2}+\frac{1}{2n}\right]$ and that is $F(1/2+1/(2n)) - F(1/2-1/(2n))$.  If $F$ is continuous, that will approach $F(1/2)-F(1/2)=0$.
