The topologists sine curve is the set:
$$ T =\left\{\left(x, \sin\left(\frac{1}{x}\right)\right): x \in (0,1) \right\} \cup\{(0,y):y\in \mathbb{R}\}$$
A common (and fun!) exercise is to prove that the set is connected. The usual answer goes like this:
(1) the graph of the function $f(x) = \sin\left(\frac{1}{x}\right)$ is the graph of a continuous function over a connected domain, so there is an homeomorphism between its domain and the graph, thus the graph is connected
(2) $(0,0)$ is a limit point of the graph, so if we add it to the set connectedness is preserved
(3) $T$ is then the union of the set constructed in the last step and the $y$ axis, being the union of two non disjoint connected sets, $T$ is also connected.
I am interested in solving the problem directly at step 1 by finding a continuous map from some simple connected set to $T$.
My (wrong) candidate would be: Let $M = \{(0,y) \in \mathbb{R}^2 : y \in \mathbb{R} \}\cup\{(x,0) \in \mathbb{R}^2 : x \in (0,1) \}$. Let $g:M\to T$ defined as:
$$g(x,y) = \begin{cases} (x,\sin\left(\frac{1}{x}\right)) & \mbox{if $0\lt x \lt 1$,}\\\ (x,y) & \mbox{otherwise}\end{cases} $$
The function would "transform" a connected subset of the $x-y$ axis into $T$, but now I see that my construction is not continuous*. Could we change $g$ in any way to make it continuous? Or maybe there is another connected (and simple!) set for which there is a continuous map into $T$?
*we could get a sequence $(X_k,0)$ in $M$ going to $(0,0)$ but for which $g(X_k,0)$ is not going to $g(0,0)$
** I have no definition for "simple", I hope it makes sense
No. It's not possible to get a continuous surjection from your $M$ to $T$, since $M$ is path-connected and $T$ is not. More generally, any space with a continuous surjection to $T$ will have to not be path-connected. Since $T$ is one of the most basic examples of a connected space that is not path-connected, I highly doubt there is any connected space with a continuous surjection to $T$ which you would consider "simpler".