I've easily shown that Topologist's Sine Curve(TSC) is connected.Now I'm trying to fill the gaps in the following proof of "showing TSC is not path-connected" .While going through this proof i got some questions,i was unable to answer .
Suppose there is a path $f:[a,c]\rightarrow \bar S,$ from $(0,0)$ to (1,0) in the Topologist's Sine Curve(TSC).
Consider the set of those $t$ for which $f(t)\in 0\times [-1,1]$ is closed,let this set be $K$.So,K is closed(K contains all of its limit points).So, K has a largest element 'b'(say) ({1}).( check the arguments in this paragraph,if there is some scope of improvement please let me know )
Then $f:$[b,$c$]$\rightarrow \bar S$ is a path that maps b into the vertical interval {$0$}$\times [-1,1]$ and maps other points of [b,c] to points of $S$.({2},{3})
We relabel the domain as $[0,1]$ for convenience.
let $f(t)=(x(t),y(t))$, where $y(t)=\sin({\frac{1}{x(t)}})$
As $f$ is continuous,the coordinates $x(t)$ and $y(t)$ are continuous as well.
We produce a sequence $<t_n>\rightarrow 0$ in $[0,1]$ such that $y(t_n)\not\to y(0)$,which violates the sequential criteria for convergence in Metric spaces.
Since,$f:[0,1]\rightarrow \bar S$ is a path.
So,$f(0)=x(0)$ and $f(t)=y(t)$ for $t>0$.
But,$f(0)=0$,So,$x(0)=0$ and $f(t)=y(t)$ for $t>0$.
Now,$x(0)=0$ and $y(t)=\sin (\frac{1}{x(t)})$ for $t>0$.
To find $<t_n>$,we proceed as follows:
Given $n$,choose $u$ with $0<u<x(\frac{1}{n})$ such that $\sin(\frac{1}{u})=(-1)^n$.
Since,$x(t)$ is continuous on $[0,1]$.So, by intermediate value theorem there exists $t_n\in (0,\frac{1}{n})$ such that $x(t_n)=u$ i.e.,$\sin(\frac{1}{u})=\sin(\frac{1}{x(t_n)})=y(t_n)=(-1)^n.$({4})
Collection of queries from the above proof-
$1.$ Is "b" largest limit point of $K$?
$2.$ How can a single point b be mapped to $0\times [-1,1]$?
$3.$ How we're so sure that it is only the largest element b which is mapped on the vertical line?
$4.$ What is explicit representation of $t_n$?
Reference:Topology by J.R.Munkres(2nd ed.)
As remarked by Cave Johnson, $K$ is not a subset of $L = \{ 0 \} \times [-1,1]$. In fact $K = f^{-1}(L) \subset [a,b]$. It is closed in $[a,b]$ because $f$ is continuous and $L$ is closed in $\overline{S}$. Hence $K$ is compact.
You correctly define $b$ as the maximum of $K$ (which is of course unique) and consider the restriction $f : [b,c] \to \overline{S}$. But then $f(b) \in L$ because $b \in K$ and $f(t) \in S$ for $t > b$ because $t \notin K$. Of course, all $t \in K \setminus \{ b \}$ are also mapped by $f$ into $L$, but this does not play any role.
The rest of your proof is absolutely correct. However, the $t_n$ cannot be explicitly represented unless you have an explict formula for $f$.