Path connected sets (How to intutively find which set is path connected before attempting a formal proof ) I have been preparing for an objective masters exam and  general topology is in the syllabus. The instructor in undergrad course was not interested in teaching and so I self studied and managed to do average level understanding of topology.
Problem arises in path connected set . Proving a set  path connected by definition is not easy and questions are often asked in exam whether a set is path connected or not? But rigorious proof is not asked as I have to just mark the correct options. So, I am asking for if there is some intution .  Consider that  I am given a set and  I have to see whether it's path connected or not , then I can use that intutive method which is less rigorious and time saving also.
If yes , can you tell me by using it on this question : Let A be the following subset of $\mathbb{R}^2$ : A={$(x,y)$:${(x+1)}^2+y^2 \leq  1$} $\bigcup $ {$(x,y): y = x sin(1/x), x>0$}. (This was 1 of the question asked in that exam).
I will apply your method/explanation/intution to other asked question for practice.
I shall be really thankful for any help received!
 A: The crucial step is to get a clear idea of the set in question. In your example it can easily be sketched and seen to consist of a unit disk centred at $\langle-1,0\rangle$ together with what we might call a shrinking topologist’s sine curve. After that it really depends entirely on the set in question; I can’t think of any general principles that are genuinely helpful. In this case, however, it’s not hard.
The origin stands out, because it’s the one point where the disk and the curve meet. If the disk and the curve are both path connected, the whole set should be: for any $p$ and $q$ in the set we should be able to get from $p$ to $q$ by concatenating a path from $p$ to the origin with one from the origin to $q$. (This is inefficient if $p$ and $q$ are on the same side of the origin, but it still works.)
It’s intuitively clear that there is a path from any point of the disk to the origin, since the line segment between the two points lies entirely within the disk. (It wouldn’t even be hard to write the equation of such a path.) And the shrinking sine curve itself provides a path from any point on the right to the origin, so the whole set is indeed path connected.
A: Here's a very vague intuition for it: Being connected means that you can't pull apart the space into several pieces. Being path-connected means that you can actually travel from any one point to any other.
You can see this quite easily with the topologist's sine curve; the two path-connected components can't be pulled apart without changing the topology, but there's no way to move continuously from one to the other. Similarly, with the one-point compactification of the long line, the point at the end can't be pulled away, but it's "too far away" to actually reach from anywhere else on the long line.
It's worth mentioning that manifolds are connected if and only if they are path connected. So any distinguishing example is going to be somewhat not intuitive.
A: Let $E = \bigg\{ (x,y)\bigg|y=x\sin\ \frac{1}{x},\ 0<x\leq 1\bigg\}$. We have a claim that $E'=\{(0,0) \} \bigcup E$ is path-connected,
i.e. we must construct a continuous map $c : [0,1] \rightarrow
(\mathbb{R}^2,|\ |)$ where $|\ |$ is Euclidean distance s.t.
$$c(0)=(0,0),\ c(1) =(1,\sin\ 1)$$ and $c([0,1])$ is in $ E'$.
Define $c_n : [0,1]
 \rightarrow
(\mathbb{R}^2,|\ |)$ s.t. $c_n(t)=(t,t\sin\ \frac{1}{t}),\
\frac{1}{n}\leq t$, $$ c_n(s) =  \frac{s}{\frac{1}{2n}}
\bigg(\frac{1}{n},0\bigg) ,\ 0\leq s\leq \frac{1}{2n} $$ and $$ c_n(s) =
\frac{ \frac{1}{n}  -s }{\frac{1}{2n} }  \bigg(\frac{1}{n},0\bigg)
 + \frac{s- \frac{1}{2n}}{\frac{1}{2n}} \bigg(\frac{1}{n},\frac{1}{n}\sin\ n\bigg),\ \frac{1}{2n}<s<\frac{1}{n}
$$
Define $c(0)=(0,0),\ c(t)=(t,t\sin\ \frac{1}{t})$ so that
$|c_n(t)-c(t)| \leq  \frac{2\sqrt{2}}{n}
 $ for all $t$.
Hence $c$ is a uniform limit of continuous curves
$c_n$. That is $c$ is continuous.
