In Munkres (Chapter 3, Section 23, p. 148), Munkres shows that if a subspace $Y$ of a space $X$ is not connected, then there are two disjoint open subsets $A,B$ such that the union of $A$ and $B$ contains $Y$. What doesn't make sense is that a separation of $Y$ only requires two open subsets of $Y$ which are disjoint, *as subsets of $Y$*. That is, $A$ and $B$ certainly can't intersect anywhere in $Y$, but who says they have to be disjoint? Or is asking whether a subset $Y$ of $X$ is connected *as a subset of $X$* a different question than saying "okay, here is the topology on $Y$ as a subset of $X$ - now is $Y$ connected under that topology?"



Definition of Connected Subspace
Per your comment, I think you misunderstood what Munkres is trying to say. 



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> If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y.
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I read it to mean two definitions are given. Firstly, that he defines what it means to have a separation of a subspace $Y$ inside of the space $X$. Then he defines a space $Y$ as connected if it cannot be separated within itself. ($Y$ is trivially a subspace of itself, so the first definition of a separation can be used.)


Now, in the example I gave where $X = \{a,b,c\}$, $Y = \{a,c\}$, with topology on $X$ generated by $\{a,b\}, \{b,c\}$, the subspace $Y$ is not a connected set in $X$, as it is not a connected space in its subspace topology. But the space $X$ is connected, so the connected component of $\{a\}$ in $X$ is the whole space. (Whereas the connected component in $Y$ is itself.)


This shouldn't be so strange if you consider a more intuitive example: Let $X$ be the open interval $(0,1)$, and $Y$ be the subset $(0,1/4)\cup (3/4,1)$. Then $Y$ is not connected. The connected component containing the point $1/8$ in $X$ is the whole space, whereas the connected component when considered in $Y$ is just the interval $(0,1/4)$. 


In other words, the connected component of a point in $X$ is a subspace $Y$ such that $Y$ is connected in the subspace topology and such that $Y$ and $X\setminus Y$ are both open. (And $Y$ of course contains the point in question.)


