Understanding the following proof about path-disconnectedness of topologist sine curve. Given the function $f$ from $\Bbb R^+_0$ to $\Bbb R$ defined by the equation
$$
f(x):=\begin{cases}\sin\Big(\frac 1 x\Big),\,\text{if }x\in\Bbb R^+\\0,\,\text{if }x=0\end{cases}
$$
for any $x\in\Bbb R^+_0$ then we call Topologist sine curve the graph of $f$ that is the set
$$
\mathcal G_f:=\biggl\{x\in\Bbb R^2:x_1\ge 0\wedge x_2=\sin\Big(\frac 1 {x_1}\Big)\biggl\}\cup\big\{\mathbf 0\big\}
$$
Now it is a well know result that $\cal G_f$ is connected but not path-connected and thus I would like to discuss this about the path-disconnectedness of $\cal G_f$: so I do not really understand why it is possible I have to remove a part of $[0,1]$ and thus rescaling; then I do not understand why the path-disconnectedness of $\cal G_f$ follows by showing that
$$
\sup\Big\{t\in[0,1]:f\big([0,t]\big)=\mathbf 0\Big\}=0
$$
finally I would like to see a formal proof about the subjectivity of $\sin\Big(\frac 1 x\Big)$ on $(0,\varepsilon)$ onto $[-1,1]$.
Please do not closed this question: unfortunately the users who wrote the linked answer and question are inactive by many time so that I cannot use comments to ask clarifications but the linked answer seems very simple and unfortuately I did not understand other proofs about the path-disconnectedness of $\cal G_f$ so that I really like to understand the linked proof: so could somoene help me, please?
 A: 
I do not understand why the path-disconnectedness of $\mathcal{G}_f$ follows by showing that: $$\sup_{t\in[0,1]}\{f([0,t])=\{0,0\}\}=0$$

This is a misinterpretation of the linked answer. They don’t show that, indeed they suppose that by a “without loss of generality” argument.
Let me call the sine curve $T$ and the first and second components of a function into $T$ by subscript $1,2$ respectively. For such a function, $f_2=\sin(1/f_1)$ is forced (when $f_1\neq0$). Suppose $f$ is any path in $T$ joining $\{0,0\}$ to some other anonymous distinct point. It is impossible for the above supremum to equal $1$ (why?). Furthermore $f(0)=\{0,0\}$, and a point implicitly used later in their answer is that if $f_1(t)=0$ then $f_2(t)=0$ is forced. If $t_0\gt0$ is the supremum, by continuity $f(t_0)=\{0,0\}$. Then a rescaling of the function $f$ to the continuous path (with the same endpoints): $$g:[0,1]\to T, \,t\mapsto\begin{cases}f(t_0+t(1-t_0))&t\in(0,1]\\\{0,0\}&t=0\end{cases}$$Would set the supremum to $0$ and moreover not change the validity of the following argument.
The argument can be restated as follows. We take without loss of generality this modified path $g$ which has the supremum of $0$ and by continuity $g(t)\neq\{0,0\}$ for $t\gt0$. Continuously pullback the unit ball about the origin by $g$, yielding an interval $[0,t’)$  such that $|g_2|\lt1$ on this interval. This is the problematic step: try visualising this, it is impossible for any path touching $\{0,0\}$ to avoid $1$ continuously. Now by the intermediate value theorem $g_1$ attains an interval of nonzero values in $g_1[0,t’)$ (by the without-loss-of-generality argument and the italicised remark $g_1[0,t’)$ is a nondegenerate interval), so by the surjectivity of the reciprocal sine curve (consider $\frac{2}{n\pi},n\in\Bbb Z$, will eventually fall into any interval about $0$) it must be the case that $g_2([0,t’))=[-1,1]$, a contradiction (recall $g_2=\sin(1/g_1)$).
I hope that clarifies things, feel free to query further. Note that there is also nothing special about the number $1$ here; it is just aesthetic.
