Let $(X,d)$ be an unbounded connected metric space. Let $x \in X$ and $r>0$ be arbitrary then there exists $y \in X$ such that $d(x,y) = r$. Let $(X,d)$ be an unbounded connected metric space. Let $x \in X$ and $r>0$ be arbitrary then there exists $y \in X$ such that $d(x,y) = r$.
We assume on the contrary that there does not exist such points $y \in X$ such that $d(x,y) = r$. Then we take the closed ball $B[x,r]$ and by our assumption the boundary of the ball does not exist in $X$, because if one point on the boundary exist then we can choose the point as our $y$ and we are done.
Now since $(X,d)$ be an unbounded connected metric space we can find points $y$ in $X$ such that $d(x,y) > r$ and $d(x,y) < r$ and we consider them as two disjoint non-empty subsets such that $X =A \cup B$ and thus contradicting $(X,d)$ is connected.
Is the logic of the proof correct? It is intuitive proof I have thought, but how can we write a good proof?
 A: The first part of your proof seems to be superfluous. Here is how I would prove this: Since $X$ is unbounded, the set $$U=\{y\in X:d(x,y)>r\}$$ is nonempty. Since $d(x,x)=0$, the set $$V=\{y\in X:d(x,y)<r\}$$ is nonempty. It is straightforward to show that $U$ and $V$ are open. Since $X$ is connected, we cannot write $X=U\cup V$, as that would be a separation of $X$. So there must be a point $y\in (U\cup V)^c$, that is, a point $y\in X$ with $d(x,y)=r$.
A: For another proof, fix $x \in X$. Note that the map $f: X \to \mathbb{R}$ given by $y \mapsto d(x,y)$ is continuous. This is true because if $d(y,y')<\epsilon$, then $|f(y)-f(y')|=\big|d(x,y)-d(x,y')\big| \leq d(y,y') < \epsilon$.
Moreover, the map $f$ is unbounded, since $X$ is unbounded. Indeed, if $M>0$, then we can find $a,b \in X$ such that $2M<d(a,b) \leq f(a)+f(b)$, and then $f(a)>M$ or $ f(b)>M$.
Since $X$ is connected and $f$ is an unbounded continuous function on $X$ you can conclude by the intermediate value property that for any $r>0$, there exists $y \in X$ such that $f(y)=d(x,y)=r$.
