If a set $E$ is connected, is every interior point of $E$ a limit point of $E$? Upon reviewing Rudin's basic topology section, I was asking myself:"Is every interior point limit point?" Because at first it seems like an interior point $p$ of a set $E$ always contain something in $E$ that is not $p$ itself.
But then I think if the discrete finite set $E$, for which each point is an interior point but not limit point because it is a finite set.
Then I thought what if $E$ is connected, not discrete -- does this mean $E$ is not finite? Then does this mean every interior point is actually a limit point of $E$?
Comments are much appreciated! Intuition tells me yes, but I don't have enough techniques yet to prove. Thanks! 
 A: Yes, this holds in $T_1$ spaces, i.e. topological spaces where all singletons are closed. This is certainly true in metric spaces, as for $y \neq x$ we have that $B(y, d(x,y) ) \cap \{x\}$ is empty, showing that all $y \in X \setminus \{x\}$ are interior points. 
Firstly $T_1$ implies that all finite sets are closed (as finite unions of closed singletons). And so every finite $F$ is discrete in itself, so disconnected if it has 2 or more points. Hence connected sets are infinite.
Also if $C$ is connected and has more than one point, and if $x$ is in the interior $C^{\circ}$ (it could also be empty, of course, like a circle in the plane) then $x$ is a limit point of $C$. If not, $\{x\}$ would be open in $C$, and $C$ would not be connected (as $\{x\}$ and $C \setminus \{x\}$ would separate it). 
Of course, if $X$ has an isolated point, $\{x\}$ would be connected and $x$ would be in its interior and no limit point, so we do need more than one point in $C$.
A: In any metric spaces $(X,d)$ (which is what Rudin dealt with mostly in his PoMA), every interior points of a connected set is a limit point. To see why, observe that for $x\in C^o$, where $C$ is a connected subset of $X$, there exist $r>0$ such that $x\in B(x,r)\subset C$.
For any $D\in \Bbb R$ such that $0<D<r$, there exist $y_D\in B(x,r)$ such that $d(x,y_D)=D$ or else $C$ can be "disconnected" by 
$$ C = \{y\in C|d(x,y)<D\}\cup \{y\in C|d(x,y)>D\}\ $$
By letting $D\to 0$ we have $y_D\to x$ so $x$ is indeed a limit point of $C$.
Edited: As Mr. Fischer has pointed out, the connected set need to consist of more than 1 element in order for the argument to work. If $x\in X$ is an isolated point then the singleton $\{x\}$ is a connected set which is, obviously, not a limit point. 
