How can I prove this function is not continuous for every point other than 0? Define $g:[0,1]\rightarrow\mathbb R$ by $g(x)=\sqrt{x}$ if $x$ is rational and $g(x)=0$ if x is irrational. 
Prove that $g$ is continuous at $x=0$, but is not continuous at any other value of $x$.
I am really at a loss of where to even start my proof. I know that in order for $g$ to be continuous as any point (let's say point $p$) then for every $\varepsilon>0$ where $E\subset X$ there exists a $\delta>0$ such that $d_Y(g(x),g(p))<\varepsilon$ for all points $x\in E$ for which $d_X(x,p)<\delta$
I'm having trouble applying this definition.
 A: a) $g(x)$ is continuous at $x=0$:
Proof:
Let $x_n \in [0,1]$ such that $x_n \to 0$, then $g(x_n) = 0$ if $x_n$ is irrational and $g(x_n) = \sqrt{x_n} \leq x_n$ for rational $x_n \in [0, 1]$.  Hence, $\lim_{n \to \infty} g(x_n) = 0 = g(0)$ for any sequence $x_n \to 0$, so $g$ is continuous at $x = 0$.  
b) $g(x)$ is not continuous at any other point in $[0, 1]$.  
Let $x \in (0,1]$.  Then if $x$ is rational, $g(x) = \sqrt{x} > 0$.  Now just choose a sequence of irrational number $y_n \to x$, then $\lim g(y_n) = 0 < g(x)$.  A similar argument works if $x$ is irrational. 
A: TO prove $g(x)$ is discontinuous at every point in $[0,1]$ except at $x=0$, consider the function
$$ f(x) = \frac{1}{\sqrt{x}} $$
Notice $f$ is continuous at every point except at $x = 0$. Suppose for contradiction that $g$ is continuous at every point $[0,1]$ except at $x=0$, then the function
$$ \frac{g}{f} $$
must be continuous also. But
$$ \frac{g}{f} = \left\{
     \begin{array}{lr}
       1 & : x \in \mathbb{Q}\\
       0 & : x \notin \mathbb{Q}
     \end{array}
   \right. $$
which we know is discontinuous everywhere. Contradiction!
