Path connectedness, continuous function 
Show for the spaces $Y:=\{(x,\sin(\frac{\pi}{x})|0<x\leq 1\}$ and $X:=Y\cup(\{0\}\times[-1,1])$:
a) There does not exist a continuous function
$w:[a,b]\to X$ with $a<b$, $w((a,b])\subseteq Y$ and $w(a)\in\{0\}\times [-1,1]$
b) $X$ is not path connected

Hello,
I have a question to this task. For a) a hint is given:
Find $t^{\pm}_n\in(a,b]$ with $t^{\pm}_n\to a$ and $w(t^{\pm}_n)\to (0,\pm 1)$.
Therefore, I suppose I have a continuous function $w$ with the given properties and want to create a contradiction, with these sequences $t^{\pm}_n$.
But I need to find a function $w$ for which the properties hold first, or I can't show that $w(t^{\pm}_n)\to (0,\pm 1)$.
I guess the searched function might looks similar to $(x,\sin(\frac{\pi}{x}))$.
I also have to get, that $w(a)\in\{0\}\times[-1,1]$.
Or do I not need to give a suitable function $w$ first?
To b)
I think I can remember a similar question, where you could assess the "way" of $\sin(\frac{1}{x})$ by a divergent series.
I know how I could do it here, but would that be sufficient, to show that $X$ is not path connected?
Thanks in advance for your tips and hints.
 A: Personally I've never been a fan of hints and for this particular question I think that sequences will just make it more difficult( although I might be wrong). The idea is to prove that if $w$ is a function satisfying the properties in $(a)$, then $w$ is not continuous.
Perhaps the easiest way to see this is assuming that $[a,b]=[0,1]$ and then assuming that $w$ is continuous and satisfies the properties in $(a)$.
Let us write $w(x)=(f(x),g(x))$. A standard result in topology shows that $f$ and $g$ are continuous if and only if $w$ is continuous. (I'll leave this to you, but you can use that $w$ is a function into the product $[0,1]^2$. 
We will show that $g$ is not continuous at $0$. It's clear that $w(0)=(0,z)$ for some $z$.
For each $0<\varepsilon <1$, we must have that $w(\varepsilon)=(f(\varepsilon),g(\varepsilon)) \in Y$ and so $f(\varepsilon)>0$. 
Then as $f(0)=0 < f(\varepsilon)$ and the continuous image of a connected set is connected, we have that for each $x' \in (0,f(\varepsilon))$ there is an $x\in (0,\varepsilon)$ such that $f(x)=x'$.  
Now, assume that $g(x)\leq 0$. In particular, since $\sin(\frac{\pi}{x})$ oscillates between $0$ and $1$ for small values of $x$, we might as well pick $x'$ at a value for which $\sin(\frac{\pi}{x})=1$. In particular, we can do this for each $\delta >0$ which is small enough. 
Hence to that end, there is some integer $N$ such that $\frac{2}{4N+1}<f(\varepsilon)$.
Consequently, there is an $x< \delta$ for each small enough $\delta$ such that $g(x)=1$ and $f(x)=\frac{2}{4N+1}$
Therefore, since $g(0)\leq 0$ and $x < \varepsilon$, i.e., $|x-0|\leq \varepsilon$, we must have $|g(x)-g(0)|< \varepsilon$.
However, $$|g(x)-g(x)|=g(x)-g(0) \geq 1 \geq \varepsilon$$
This contradicts the continuity of $g$ which implies that $w$ is not continuous.
The case for $g(x)>0$ is similar by choosing a value where $\sin(\frac{\pi}{x})=-1$. 
Of course you may construct a sequence in this manner, but I think this is a little simpler.
For (b), you now have, using $(a)$ of course, that there is no path from say $(0,0)$ to $(1,0)$ or in fact from any point in $Y$ to any point in $Y\setminus X$.
