Two tangent closed discs connected 
Let $X$ be a connected subset of a metric space $M$. Show that $X^0$ (the interior of $X$) is not necessarily connected.

So the example I'm thinking of is $X$ being two closed discs, tangent at a point. The interior is clearly not connected, since there exist two disjoint, non-empty open sets whose union is the interior (the two sets are exactly the two discs.) 
But I don't know how I can prove that $X$ is connected. Using the definition of connectedness, I must prove either
(i) The only subsets of $X$ which are both open and closed are $\emptyset$ and $X$
or
(ii) If $A,B$ are disjoint open subsets of $X$ whose union is $X$, one of them contains $X$.
 A: Connectedness can be a bit of an abstruse concept to work with. It's often easier to work with the stronger concept of path-connectedness (a space is path-connected if any two points can be joined by a continuous path in the space). Not every connected space is path-connected, but for those that are, this is generally the easiest way to prove connectedness.
In this case, for example, it's almost trivial to see that $X$ is path-connected: two points in the same disc can be joined by a straight-line path, while two points in different discs can be joined by a composite path formed of two line segments meeting at the tangency point.
A: This is a bit of thread necromancy here, and you may have well already learned this in the time between the original post of this question and my response here, but your question can be generalized a good bit without much work.
Particularly, if we have two connected subspaces $X, Y\subset Z$ such that $X$ and $Y$ aren't disjoint, then $X\cup Y$ is connected (this obviously applies to your case since $X=Y=D^1$ are tangent). Say we have a separation of $Z$, i.e. disjoint open sets $U, V$ whose union contains $Z$. Then since $X$ is connected, either $X\subset U$ or $X\subset V$ (can you see why?) and similarly for $Y$. Moreover, since $X$ and $Y$ aren't disjoint, they need to be contained within the same set, either $U$ or $V$. But this means that $Z$ is contained in either $U$ or $V$, so $Z$ is connected. This is all in Munkres, Introduction to Topology in the chapter on connectivity.
This also works if we have an arbitrary collection of connected subspaces $\{X_i\}_{i\in I}$ which are pairwise non-disjoint.
