Page 155 of the second edition of Munkres "Topology".... I've seen a lot of proofs of this and maybe it's because it's 2 a.m. but something isn't clicking for me atm.
Suppose there is a path $f: [a,c] \rightarrow Cl(S)$ beginning at the origin and ending at a point of S. The set of those t for which $f(t) \in 0 \times [-1,1]$ is closed, so it has a largest element b. Then $f: [b,c] \rightarrow Cl(S)$ is a path that maps b into the vertical interval $0 \times [-1,1]$ and maps the other points of $[b,c]$ to points of S.
Replace $[b,c]$ by $[0,1]$ for convenience; let $f(t)=(x(t),y(t))$. Then $x(0)=0$ while $x(t)>0$ and $y(t)=sin(1/(x(t))$ for $t>0$. We show there is a sequence of points $t_n \rightarrow 0$ s.t. $y(t_n)=(-1)^n$, thus contradicting the continuity of $f$.
To find $t_n$, we proceed as follows: Given n, choose u with $0<u<x(1/n)$ such that $sin(1/u)=(-1)^n$. Then use the intermediate value theorem to find $t_n$ with $0<t_n<1/n$ such that $x(t_n)=u$
So as I typed this up something sort of clicked. The point is that under the assumption of path connectedness, that ultimately allows to invoke the intermediate value theorem with the lower bound of the range being zero, thus enabling us to find a sequence of $t_n \rightarrow 0$ and the $y$ component of $f$ not converge. Is this thinking correct? My professors proof was way different, he doesn't like Munkre's proofs of things for some reason, perhaps it has something to do with the fact he's 300 years old.
On a side note, the way he defines connectedness is that a space $X$ is connected if there is no onto continuous map from $X$ to $\{0,1\}$ with the discrete topology. It's apparent that these two definitions are equivalent but it is strange that Munkres doesn't even mention it, according to my professor it's the standard definition in literature.
Yes, the $t_n$ follow by applying the IVT to the continuous function $x(t)$.
And, no, having all continuous maps to $\{0,1\}$ constant is not `the standard definition'. It is just one of the many equivalences. What I learned was "not the union of two nonempty separated sets". This was the followed by "But those sets must be both open and closed so we can also say ..."