Is $X=\{(0,0),(1,0)\}\cup \bigcup_{n \neq 1}\{(x, 1/n) \mid x\in \mathbb{R}\}$ connected? Find its conmected components. 
Is $$X=\{(0,0),(1,0)\}\cup \bigcup_{n \neq 1}\{(x, 1/n) \mid x\in \mathbb{R}\}$$ connected with the usual topology on $\mathbb{R}^2$? Find its conmected components.

I have a vague idea if what this space $X$ looks like, and I don't know the answer so I thought I could try to provethat the set is not locally connected. 
I think that the points $(0,0), (1,0)$ might be problematic but I'm not sure how to take it from there. I think that if $U$ is an open set not containing, then it is disconnected since we could write it terms of unions of points $\{x,1/n_i\}$, but I'd need to prove that this sets are open, but I don't know how.
The plan is proving that the connected components of $X$ are the points $\{x,1/n\}$ and $(0,0),(1,0)$, then show that $(0,0),(1,0)$ are not open and hence $X$ is not locally connected. But I'd really need some help to point me at the right direction. Thanks in advance!
 A: Suppose we have two nonempty connected sets $A,B$ in a topological space X, and we're wondering if $A \cup B$ is connected. For the union to be connected, it suffices to consider any two open sets $U$ containing $A$, $V$ containing $B$, and show $U \cap V$ necessarily have non-empty intersection (why is this? try drawing a picture/proving this statement directly). If the union is disconnected, you need to find $U,V$ (U contains A, V contains B) such that $U \cap V = \emptyset$
Now your set $\{(x, \frac{1}{n})| x \in \mathbb{R}\}$ are each connected because they are path connected, (because they are horizontal lines) (You should really prove it's path connected if it's unclear). Hence it's connected. Clearly, the points $(0,0)$ and $(0,1)$ are connected as sets in their own right, so by the lemma it suffices to take an open set U around $(0,0)$, and an open set $V$ around $A=\{(x, \frac{1}{n})| x \in \mathbb{R}\}$ and prove $U \cap V \neq \emptyset$ to show the union of the two is connected, or find open balls $U, V$ s.t $U \cap V = \emptyset$ and this will show the union of the two isn't connected. Since the topology on $\mathbb{R}$ is generated by a basis of open balls, it's sufficient to take $U,V$ to be open balls. 
A: Each line $I_n={\Bbb R} \times \{1/n\}$, $n\geq 2$ is clearly a connected component. It is connected and it is closed and open (for the topology induced on  $X$). A connected component is a maximal connected set, i.e. an increasing union of connected sets which is not contained in any strictly larger connected set. Being in a maximal connected set is an equivalence relation between points so the special points $\{(0,0)\}$ and $\{(1,0)\}$ must be in components different from every $I_n$. So each of them is a connected component by itself (consisting of one point).
Note that $\{(0,0)\}$ being in the closure of the $I_n$'s  does not imply that it belongs to a strictly larger components. For example the connected components (in ${\Bbb R}$)  of $\{0\} \cup \{1/n: n\geq 1\}$ consists of one point sets (essentially the same exercise as the present).
