Connected but not path-connected Let $r:[0,\infty)\rightarrow\mathbb{R}$ be given by $r(t)=1+e^{-t}$. Let $S\subset \mathbb{C}$ be the image of the "spiral" curve $f:[0,\infty)\rightarrow\mathbb{C}$ given by $f(t)=r(t)e^{it}$. Then $\overline{S}$ is the union of $S$ and the unit circle. Prove that $\overline{S}$ is connected but not path-connected.
My intuitive idea is that $\lim\limits_{t\rightarrow \infty}f(t)=\lim\limits_{t\rightarrow\infty}e^{it}$ does not exist, but does trace out the unit circle indefinitely. So the unit circle is a limit of the spiral curve, which means that there is no way to separate the spiral curve from the unit circle, so $\overline{S}$ has to be connected.
Because $\lim\limits_{t\rightarrow \infty}f(t)$ does not exist, then any point on the spiral curve is can not "travel" to the unit circle because the spiral curve does not intersect the unit circle and any point on the unit circle can not "travel" to the spiral curve because the unit circle does not intersect the spiral curve.
Is my idea far away from the correct idea? If not, how do I make my idea into a precise, rigorous proof?
Image of the Spiral Curve
 A: Your ideas are correct, but going from there to a rigorous proof is non-trivial. I will give some hints.  It is a general fact that if $A$ is a connected subset of a topological space, then $\overline{A}$ is also connected (this may be citable, but it should not be hard to fill in the details). Also, the continuous image of a connected set is connected (also probably citable but easy to fill in the details). From this it follows that $S$ is connected and so is $\overline{S}$. 
Note that $1$ and $2$ are both on $\overline{S}$. The point $1$ is on the unit circle and $2=f(0)$.  Suppose that $p:[0,1]\to \overline{S}$ is a continuous path such that $p(0)=2$ and $p(1)=1$.   Then  by the continuity of $p$ and $|p|$, $\{x\in [0,1]: |p(x)|=1\}$ is a closed, non-empty set not containing $0$.  By closedness of this set, it has a minimum, call it $\varpi$. Since $|p(0)|=2\neq 0$, $0<\varpi$.  Then we want to consider the behavior of $p(x)$ for $x$ very close to $\varpi$ and to the left of $\varpi$.  More precisely, there exists $0<\varepsilon<\varpi$ such that if $\varpi-\varepsilon<x<\varpi$, $|p(\varpi)-p(x)|<1$.  Pick $x_0\in (\varpi-\varepsilon, \varpi)$ and note that $|p(x_0)|>1$. This means $p(x_0)$ is on $S$ somewhere. Now the only way to get $|p(x)|$ from $|p(x_0)|$ down to $1$ (which must happen between $x_0$ and $\varpi$) is to travel another loop around the unit circle. But this can't be done while maintaining $|p(\varpi)-p(x)|<1$. 
A: $S$ is the image of a continuous function from a connected set and so is connected; then, $\overline S$ is connected, too. 
For the second part, I think we can mimic the proof that the topologist’s sine curve is not path-connected:
Suppose there is a path $\phi: [0,1]$ such that $\phi(0)=(2,0)$ and $\phi(1)=(1,0).$ The first point is in $S$ and latter point is in $\overline S\setminus S.$
Let $s_0\in [0,1]$ be the least number $s$ in $[0,1]$ such that $|\phi(s)|=1.$ Of course, $|\phi(s)|\ge 1$ and  $s_0>0.$ 
The continuity of $\phi$ gives us an interval $I=(s_0-\delta, s_0)$ such that $s\in I\Rightarrow |\phi(s)-\phi(s_0)|<1/2.$ Choose any $s\in I$ and note that $3/2>|\phi(s)|>1$ so $\phi(s)\in S$ itself and so is of the form $r(t)e^{it}$ for some $t\in [0,\infty).$
So for each $s\in I,\ \phi(s)\in S$ and so $\phi(s)= r(t_s)e^{it_s}$ and $|\phi(s)|=r(t_s)$ for some $t_s\in [0,\infty).$
Now, as $s\to s_0,\ |\phi(s)|\to |\phi(s_0)|=1$ so $t_s\to \infty$ because in this case, $|\phi(s)|=r(t_s)=1+e^{-t_s}\to 1$.
So, if we choose $(s_n)$ so that the corresponding $t_n$ are $t_n= \pi/2+2\pi n,$ we get $|\phi(s_n)-\phi(s_0)|=|(1+e^{-n\pi/2})i-1|>1/2$ if $n$ is large enough, which is a  contradiction.
