The post Can a nowhere continuous function have a connected graph? asks exactly that question. The OP was interested in a few cases, one of which was when $f: \mathbb{R} \to \mathbb{R}$ under the usual topology. I'm interested in continuing to investigate the $f: \mathbb{R} \to \mathbb{R}$ case, namely can we construct a function $f$ that is nowhere continuous but whose graph is connected in the plane?
A good starting point is the topologist's sine curve
$$t(x)= \begin{cases}\sin\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \\ \end{cases}$$
and then to take an infinite sum of these with the modification that each summand is discontinuous at each rational and to scale it so that it does not grow without bound.
That is, if $\{q_n\}_{n \in \mathbb{N}}$ is an enumeration of the rationals, consider the function
$$f(x) = \sum_{n=1}^\infty \frac{\sin\left( \frac{1}{x-q_n} \right)}{n^2} := \sum_{n=1}^\infty f_n(x)$$
It's easy to see that $f$ is discontinuous at at each rational and the limit of the partial sums converges nicely. My intuition suspects that because each $f_n$ has a connected graph in the plane, then the countable sum of these should(?) have a connected graph as well. Further, due to the density of $\mathbb{Q}$, one suspects that this leads to a discontinuity at every point.
If this is not a good candidate function for this question, where does it go wrong $-$ and even more $-$ what would be a better construction?
EDIT: I'm now realizing that technically for the graph to be connected, I need to define the graph at each new iteration of $f_n$. So, for the time being, spare me that detail and let us suppose at each rational, the function is defined in a similar manner to how $0$ is mapped to $0$ in $\sin(1/x)$.
