A question about path connected sets in Topology This particular question was asked in my Topology quiz and I was unable to solve it and so I am asking for help here .

Question: Consider the following Subsets of $\mathbb{R}^2 : X_1 ={(x, sin(1/x) |0<x<1}, X_2 =[0,1]\times {0} , X_3 =${(0,1)} .  Then ,


*

*$X_1 \cup X_2\cup X_3 $ is  connected ;


*$X_1 \cup X_2 \cup X_3 $ is path connected;


*$ X_1 \cup X_2 \cup X_3 $ is not path connected , but $X_1 \cup X_2 $ is path connected ;


*$X_1\cup X_2$ is not path connected , but every open neighbourhood of a point in this set contains a smaller open neighbourhood which is path connected .
I have proved A to be true . I can visualize how the Diagram $X_1\cup X_2\cup X_3$ looks and intutively I think that it's path connected . ( Answer of question says that I am wrong )
But what difference does adding $X_3$ to $X_1 \cup X_2$ makes ? ( So , I am confused on option 2,3,4)
Actually , topology course was taught at my university but it was taken by a really terrible instructor who was not interested in teaching although I have self studied all of general topology from  Wayne Patty's "Foundations Of Topology" .
It is my humble request to  give a rigorious proof of the union of sets which is path connected and also give hint of what difference $X_3$ is making .
Answer :

 1,3

I shall be really thankful for a detailed answer .
 A: 2,3)) Suppose to the contrary that a set $X=X_1\cup X_2\cup X_3$ is path-connected. Then there exists a continuous map $f(X)\to [0,1]$ such that $f(0)=(0,0)$ and $f(1)=(0,1)$. Put $X_-=\{(x,y)\in X:y\le 1/2\}$. Since the set $f^{-1}(X_-)$ contains $0$, it is non-empty. Since $X_-$ is a closed subset of $X$ and the map $f$ is continuous, a set $f^{-1}(X_-)$ is a closed subset of $[0,1]$, and so $f^{-1}(X_-)$ is compact. Thus a set $f^{-1}(X_-)$ contains its supremum $T$. Since $f(1)\not\in X_-$, $T<1$. It is easy to see that a set $X\setminus X_-$ splits into connected components which are the arcs of the graph and the set $X_3$. Since a set $(T,1]$ is connected, its continuous image $f((T,1])$ is connected too. Since $(0,1)\in f((T,1])\subset X\setminus X_- $, the only possibility to keep  $f((T,1])$ connected is to have $ f((T,1]=\{(0,1)\}$. By the continuity of $f$, the set  $f^{-1}(0,1)$ is closed, so it contains $T$. Thus $f(T)=(0,1)$, a contradiction with $f(T)\in X_-$.
3,4)) The set $X_1\cup X_2$ is path-connected, being a union of two intersecting path-connected sets (a segment and a graph of a continuous function on an interval).
A: Your intuition is good when it comes to connectedness. However, path connectedness is a little subtle. Think of it this way : path connectedness means that if I take two points $a$ and $b$, there is a route along which, if I travel with constant speed, I can reach the other point in finite time. What that means, is that even though two points may look connected via a way, that way which is seemingly connecting them may have "infinite" length, and since you can't traverse an infinite length in a finite amount of time going at a constant speed, you lose path connectedness in the event of every such connecting way having infinite length.
The observation that clinches non-path connectedness of $X_1 \cup X_2 \cup X_3$, is the fact that to go from a point in $X_1$ to a point in $X_3$, you need to get through those infinitely many bends in the graph of the function $\sin(\frac 1x)$ near zero. Those bends make the length of that way infinite, so that way can't be a path!
However, take $X_3$ out of the way, and every point in $X_2$ actually lies "a finite distance" from $X_1$, so there can be a path from a point in $X_1$ to a point in $X_2$.

For connectedness of $X_1 \cup X_2 \cup X_3$, observe that $X_1$ is the image of $(0,1)$ under the continuous function $y \to (y,\sin (\frac 1y))$  so is connected.  Note that $X_2$ is connected trivially as a product of connected sets. Since these have at least one point in common, it follows that $X_1 \cup X_2$ is connected. Now, let $C$ be the connected component containing $X_3$ : it is easy to see that $C$ (being an open set) contains infinitely many points from $X_1 \cup X_2$ (one is enough) so that connected component also contains $X_1 \cup X_2$. Finally, $X_1 \cup X_2 \cup X_3$ is connected.

Path connectedness is more subtle, because although I did describe the intuition , the notion of length is problematic when it comes to paths. What helps is this, though. Imagine a path from $(0,1)$ to a point in $X_1$. By continuity, we can choose a time such that this path remains very close to $(0,1)$ until that time. However, by the nature of the sine curve, you can show that the path will certainly have gone far away from $(0,1)$ as well (towards the lower side of the $\sin \frac 1x$ curve), which gives a contradiction.
For the math, let $\gamma : [0,1] \to X_3 \cup X_1$ be a continuous function such that $\gamma(0) = (0,1)$ and $\gamma(1) \neq (0,1)$. We can rescale such a path to assume that $\gamma(t) \neq (0,1)$ for all $t>0$ (Basically, if $\gamma$ stays at $(0,1)$ for some time before departing, change the path speed so that the path is traversed in the remaining time).  Now, there exists $t_0 > 0$ such that for $t < t_0$ we have $\|\gamma(t) - (0,1)\| < \frac 12$. In particular, $\gamma(t)$ cannot have second coordinate $-1$ for $t<t_0$.
Let us call the component functions of $\gamma(t)$ as $\gamma(t) = (\gamma_1(t),\gamma_2(t))$. Remember, these are also continuous. Note that $\gamma(t)$, when it reaches $X_3$, must, by virtue of belonging in $X_3$, have the property that $\gamma_2(t) = \sin \frac 1{\gamma_1(t)}$. Now, simply pick any two points $0<t_1<t_2<t$ such that $\sin \frac 1{\gamma_1(t_1)} = \sin \frac 1{\gamma_1(t_2)}$ (can be done using the fact that $\gamma_1$ is continuous), and note that $\gamma_2(t)$ must then hit the point $-1$ at some point $t_3$ between $t_1$ and $t_2$ by the nature of the sine curve. This provides a contradiction, since $t_3<t_0$ but $\gamma_2(t_3) = -1$!

$X_1 \cup X_2$ poses no problems whatsoever, due to continuity : indeed, $X_1$ being the image of the path connected $(0,1)$ under a continuous function makes it path connected, and $X_2$ is obviously path connected, so them having a point in common makes their union path connected. Note that $X_3$ may be path connected by itself, but does not have a point in common with this union, hence we can't really push such a simple argument across for the previous part.

Of course, this negates part $4$, and we are done. I have left some gaps in the proofs, I expect you to fill these. My idea is the "intuition" in the proof, and how the mathematical execution is done.
