Let $A$ be the following subset of $\Bbb R^2$:
$$A=\left\{(x,y)\in\Bbb R^2:(x+1)^2+y^2\leq 1\right\}\cup \left\{(x,y)\in\Bbb R^2: y=x\text{sin}\left(\frac{1}{x}\right),x>0\right\}$$.
Prove or disprove the set is path connected.
Definition: A space $X_\tau$ is path connected if for every pair of points $x_0,x_1\in X$, there exists a path $\alpha:[0,1]_\mathfrak{U}\to X$ with $\alpha(0)=x_0$ and $\alpha(1)=x_1$.
How to solve above problem?
First use the fact that a union of two path-connected subsets, whose intersection is nonempty, is path-connected. Then try to prove for each of them separately that they're path-connected.
The set on the left is a ball, so it's convex. That is, it's pretty easy to see that the straight line between any two points inside it is contained in it, and hence is a path between them.
The set on the right is the graph of a continuous function (well, you need to prove that it is continuous, I guess). Call this function $f$. So any two points in it are of the form $(x_1,f(x_1))$ and $(x_2,f(x_2))$, and it's easy to see the path $\varphi:[x_1,x_2]\to\Bbb{R}^2$ defined by $\varphi(t)=(t,f(t))$ is a path between them.