(a) It is reflexive and symmetric. To show transitivity suppose that $x\sim y\sim z$ but $x\not\sim z$ . There is a separation of points $x$ and $z$ , and $y$ is in one of these open sets, i.e. it is either separated from $x$ or $z$ . Contradiction.(b) If for two points in one component there is a separation of these points, then the component is not connected. On the other hand, what can make a component to be a proper subset of a quasi-component? Or, equivalently, when a quasi-component may contain two or more components? Obviously, it cannot be the case if the components are open, as otherwise they would be open and closed and form a separation for points in different components. Therefore, if the space is locally connected, then the components are open, and quasi-components are the same as components.(c) $A$ is not (locally) (path) connected. For any two points with different $x$ -coordinates $x<x'$ consider a point $x"$ such that $x<x"<x'$ and $1/x"\notin\mathbb{Z}$ . Then open sets $x<x"$ and $x>x"$ form a separation of the points. Now, each vertical line segment is path connected, therefore, it is a path component, component and quasi-component. The remaining two points are disconnected, therefore, there are two more (path) components. The only question here is whether there is one or two quasi-components. We show that there is only one quasi-component with both points in it. The idea is that any separation of the space must "draw a line" between a pair of vertical lines without intersecting any of them, and therefore, the two points must be in one set in the separation. Suppose there is a separation $A'\cup A"=A$ such that $(0,1)\in A'$ and $(0,0)\in A"$ . For each open set $A'$ and $A"$ take a small open ball-neighborhood of the corresponding point contained in the set. The balls intersect some common vertical line, therefore, $A'$ and $A"$ intersect that line and form its separation. This contradicts the fact that the line is (path) connected.$B$ is connected ($B-\{(0,1)\}$ is path connected and adding a limit point keeps it connected), so that there is one component = quasi-component. But there are two path-components: $(0,1)$ and all other points.$C$ is not locally (path) connected. Its path components are the line segments (so there are infinitely many of them). We show that it is connected and, hence, has one (quasi-)component. Suppose there is a separation $C'\cup C"=C$ . Both are open, therefore, they cannot intersect the same line segment. Moreover, they do not contain limit points of each other. Take any point $(-1/n,0)$ and suppose it is in $C'$ . It is a limit point for the set of line segments $[-1,0]\times K$ . Therefore, starting from some $k$ all lines $[-1,0]\times(1/k)$ are in $C'$ . Therefore all points $(-1/n,0)$ are in $C'$ . Therefore, all line segments $-K\times[-1,0]$ are in $C'$ . This implies all their limit points $(0,-1/n)$ in $C'$ . By continuing this argument we conclude that all points are in $C'$ , and $C"$ must be empty. Contradiction.