In general $\overline{U_\epsilon(x)}=K_\epsilon(x)$ false First of all let $(M,d)$ be a metric space. We know that the set $K_\epsilon(x)=\{y\in M \mid  d(x,y)\le\epsilon\}$ for arbitrary $x\in M$ and $\epsilon>0$ is closed and $\overline{U_\epsilon(x)}\subseteq K_\epsilon(x)$
In $\mathbb R^n$ with the Euclidean metric we have an equality but in general it is false. 
Now my question: What does this inequality has to do with $M:= S^1\cup\{(x_1,0)\in\mathbb R^2 \mid 0\le x_1\le 1\}$ with the constraint of the euclidean metric and the corresponding spheres with radius 1 ?
Here $K_\epsilon(x)=\{y\in M \mid  d(x,y)\le\epsilon\}$ and  $U_\epsilon(x)=\{y\in M \mid  d(x,y)<\epsilon\}$
 A: In your example, the closure of $[0,1)\times\{0\}$ (the open unit ball in $M$) is $[0,1]\times\{0\}$, but the closed unit ball is all of $M$
A: ADD In your case, let $$K_1(0)=\{x\in M:d(x,0)\leq 1\}$$ Then $K_1(0)$ is all of $M$. On the other hand, $$U_1(0)=\{x\in M:d(x,0)<1\}=\{(x,0):0<x<1\}$$ and the closure of $U_1(0)$ in $M$, that is, relative to $M$, is $$\overline{U_1(0)}_M=\{(x,0):0\leq x\leq 1\}$$

It seems you're wondering about relative closures in metric spaces. Suppose we're working in a metric space $(X,\rho)$. Fix some subset $E$ of $X$. Then we say $F\subseteq E$ is relatively open in $E$, or open relative to $E$, or $E$-open if $F=E\cap O$ for $O$ open in $X$. We analogously say $K$ is closed relative to $E$ if $K=E\cap L$ for some closed $L$ in $X$. 
With this in mind, we can talk about relative closures and relative interiors of sets. In particular fixing a subset $E$ of $X$, we define the closure of $F\subseteq E$ as $$\operatorname{cl}_E F=\bigcap\mathscr C$$
where $\mathscr C$ is the collection of all relatively closed sets in $E$ that contain $F$. You can then see that $$\operatorname{cl}_E F=E\cap\operatorname{cl}  F$$
We can do the same for $\operatorname{int}_E F$. Thus, in general you'll have that 
$$\operatorname{cl}_E F\subseteq \operatorname{cl}  F$$
$$\operatorname{int}_E F\supseteq \operatorname{int}  F$$
You can even do this for nested subsets $M\subseteq N\subseteq X$ for $X$ a topological space. In such case we will have for $F\subseteq M\subseteq N$
$$\operatorname{cl}_M F\subseteq \operatorname{cl}_N  F$$
$$\operatorname{int}_M F\supseteq \operatorname{int}_N  F$$
