Showing an Integrable function is everywhere discontinuous 
Question
  Let $f(x) = x^{-1/2}$ for $0<x<1$ and $0$ otherwise, $r_n$ be an
  enumeration of rationals. Define
  \begin{equation}
  g(x) = \sum_n g_n(x) \quad \text{where} \quad g_n(x) = 2^{-n}f(x-r_n)
\end{equation}
  Show that $g$ is discontinuous everywhere.

Attempt
I claim $g$ is unbounded for any interval $I$, which
  implies it cannot be continuous anywhere. Note that if $(x-r_n) \leq 2^{-2n}$
  then $g_n(x) \geq 1$. Construction relies on the fact that
  any interval contains infinitely many $r_n$.
For $I$, there exists $r_k\in I$ with a corresponding interval $I_k
  \subset I$ of size $2^{-2k}$ with left endpoint $r_k$. Repeat for
  $I_k$, there exists $r_{k'}\in I_k$ with a corresponding interval
  $I_{k'}\subset I_k$ of size $2^{-2k'}$ with left end point $r_{k'}$. 
  Therefore, there exists an increasing sequence of $r_k$'s
  and they converge, say to $x$. Note that $(x - r_k) \leq 2^{-2k}$
  for all $k$ in this construction. It follows, $g(x) = \sum_n g_n(x)
  \geq \sum_k g_k(x) \geq \sum_k 1= \infty$.
Appreciate if one can verify/point out the mistake/suggest another approach. Notation is not very precise, but hopefully explains it.

Apparently the codomain of $g$ includes $\infty$. That can in itself be okay; there are good topologies on the extended real line.
Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=\infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=\infty$. Then $g$ would be perfectly continuous in the interior of such interval.
So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)
