Connected components of subspaces vs. space If $Y$ is a subspace of $X$, and $C$ is a connected component of $Y$, then C need not be a connected component of $X$ (take for instance two disjoint open discs in $\mathbb{R}^2$).
But I read that, under the same hypothesis, $C$ need not even be connected in $X$. Could you please provide me with an example, or point me towards one?
Thank you.
SOURCE http://www.filedropper.com/manifolds2 Page 129, paragraph following formula (A.7.16).
 A: Isn't it just false? The image of a connected subspace by the injection $Y\longrightarrow X$ is connected...
A: Are you sure you read correctly? Suppose $C$ is not connected in $X$. By definition this means that there exists two open sets $U, V\subset X$ such that $U\cap C \neq \emptyset$, $V\cap C\neq \emptyset$, $U\cap V = \emptyset$, and $C\subset U\cup V$. But then by definition of subspace topology, $U' = U\cap Y$ is open in $Y$, and $V' = V\cap Y$ is open in $Y$. And since $C\subset Y$, you have that $U' \cap V' = \emptyset$ and $C \subset U'\cup V'$, and $C \subset C\cap U \subset Y\cap U = U'$ etc. so $C$ cannot be connected in $Y$. 
A: As @LostInMath pointed, the example below is wrong because in the definition of disconnected, the open sets do NOT need to be disjoint. The example below just shows that a set might be disconnected and yet there are no disjoint open sets separating them.

I guess what you want is this:
Let $C \subset Y \subset X$.
Even if $C$ is connected in $X$, it doesn't mean it is connected as a subset of $Y$. In fact, it is possible that $Y$ is connected as a subset of $X$ but disconnected as a topological space.
Let $X = \{a, b, c\}$ with the topology given by the sets
$\emptyset, X, \{a,b\}, \{b,c\}$.
And let $Y = \{a,c\}$.
No $X$ subset is disconnected since every non-empty open set has $b$ as a common element. But $Y$ is disconnected. In fact, it is discrete.
