Locally compact metric space having atleast one non-compact closed ball Can we have a non-trivial example of a locally compact metric space in which atleast one non-trivial closed ball is not compact.
I am considering that infinite set with discrete metric is a trivial example. More than one examples are appreciated. 
 A: The real numbers, usual topology, but metric $d(x,y) = \min\{1,|x-y|\}$.  It is still locally compact, but $\overline{B}(0,1)$ is the whole space and is not compact.
A: Take $(-1,1)$, endowed with the usual distance. It is locally compact and, if $r>1$ and $x\in(-1,1)$, the closed ball centered at $x$ and with radious $r$ is non-compact. Note that, it some cases, it is not the whole space either (such as when $x=\frac45$ and $r=\frac32$).
A: Let $(X,d)$ be any non-compact metric space.
Form the standard bounded metric $\bar{d}=\min\{d,1\}$ (note that $d$ and $\bar{d}$ generate the same topology).
If $r\geq 1$, the closed ball $\bar{B}(x,r)=X$ and is therefore not compact.

Note: The original question was edited to impose the condition that the closed ball also be non-trivial, so this answer now does not furnish an example.
A: Here is another example.  Take $X=(0,1)\times S$ where $\# S\geq 2$, with the distance function
$$
d((t_1,s_1),(t_2,s_2))=
\begin{cases}
|t_1-t_2| & s_1=s_2\\
2 & s_1\neq s_2
\end{cases}
$$
i.e., you are taking $\#S$ disjoint copies of the interval and declaring the distance between two points to be the same as that on $(0,1)$ if they are on the same copy and $2$ otherwise (there is nothing special about $2$, any $M>1$ works).
Then $\overline{B}_1((t,s))=(0,1)\times\{s\}$ is non-compact and not the whole space $X$.
Note that you could replace $(0,1)$ by any noncompact, locally compact metric space with diameter $\leq 1$.
