I'd like to clarify a few points about the example below. Here are my questions:
1) The set of $t$ for which $f(t)\in 0\times[-1\times 1]$ is closed because it's the preimage of the closed set $0\times [-1,1]$ under a path?
2) Why does the above set have a largest element $b$? Is it because any closed subset of $[a,c]$ is of the form $[r,s]$ for some $r>a,s<c, r<s$? (In this example $s$ is a largest element, right?)
3) Why is it possible to choose $u$ between $0$ and $x(1/n)$ in such a way that $sin(1/u)=(-1)^n$?
4) How exactly is the Intermediate value theorem applied at the end?
1) Yes.
2) Because the set is a closed set of the compact set $[a,c]$, hence is itself a compact set. The compact sets of a closed interval are its the closed bounded sets. It can also be of the form $[r,s]\cup[u,v]$ and so on.
3) Let $n\in\mathbb N$. $\sin$ is a periodic function with values in $[-1,1]$. Thus the set of points at which the value of $\sin$ is $(-1)^n$ is unbounded. So, you can find $v$ as big as you wish such that $\sin(v)=(-1)^n$. So you can choose $v>\dfrac{1}{x(\frac{1}{n})}$. Now choose $u=\dfrac{1}{v}$.
4) $x$ is continuous and $0=x(0)<u<x(\frac{1}{n})$. By the intermediate value theorem, you can find $t_n\in (0,\frac{1}{n})$ such that $x(t_n)=u$. Notice that $(-1)^n=\sin(\frac{1}{u})=\sin\left(\frac{1}{x(t_n)}\right)=y(t_n)$.
This constructs a sequence $(t_n)_{n\in\mathbb N}$ that converges to $0$ (because $0<t_n<\dfrac{1}{n}$) and such that $x(t_n)>0$ and $y(t_n)=(-1)^n$. Munkres goes through $x(t_n)=u$ to have $x(t_n)>0$.