The discret distance on a set $E$ is given by $d(x,y)=\begin{cases} 1,~x\neq y\\ 0,~ x=y\end{cases}$ and let an other distance $d'$ on $E$.
To prove that $d$ is thiner than $d'$ we prove that any open A related to $d'$ is open related to $d$.
$A$ is an open related to $d'$ iff $\forall x\in A, \exists r>0, B_d'(x,r)\subset A$
$B_d(x,r)=\begin{cases}\{x\},~ 0<r<1,\\ E,~ r\geq 1\end{cases}$
If $0<r<1$ then we are done $B_d(x,r)\subset B_{d'}(x,r)\subset A$, but how to do if $r\geq 1$?
Thank you
That's a "there exists" quantifier. You've shown that $r=\frac12$ works. That's enough; you're done, and you don't need to worry about any other $r$. You have a ball centered at $x$ inside $A$, which is what you were looking for.
The discrete metric induces the discrete topology, in which every set is open.