Topologicaly equivalent using balls We have two metrics, $d$ and $d'$ on an open set $\Omega$, such that $d'(x,y)=\sup\left(d(x,y),\left|\frac{1}{f(x)}-\frac{1}{f(y)}\right|\right)$ and $f(x)=d(x,\partial\Omega)$. 
How can we prove that $d$ and $d'$ are topologically on $\Omega$ without using sequences?
If we take $y\in B_d'(x_0,r)$, then $d'(y,x_0)<r$, then $d(y,x_0)<r$: so $y\in B_{d}(x_0,r)$. 
But if we take $y\in B(x_0,r)$, then $d(x_0,y)<r$. How do we find that $d'(x_0,y)<r$?
 A: If you want to show the equivalence of the two metrics $d$ and $d'$ on $\Omega$, you need to show two things:
$$\text{1. } \forall x \in \Omega: \forall r>0: \exists s>0 : B_{d}(x,s) \subseteq B_{d'}(x,r)$$
(Which shows that all $d'$-open sets are $d$-open) and
$$\text{2. } \forall x \in \Omega: \forall r>0: \exists s>0 : B_{d'}(x,s) \subseteq B_{d}(x,r)$$
(Which shows that all $d$-open sets are $d'$-open)
In this case, $d \le d'$ so 2. is trivial: we can take $s=r$,as you indicated.
For the other you have to use the continuity of $\frac{1}{f(x)}$ on $\Omega$. Here you cannot use the same $s=r$, as you seem to think, though.
Added upon request: Let's take $(X,d)$ a metric space and let $h: (X,d) \rightarrow \mathbb{R}^+$ be a continuous function, where the latter set has the usual Euclidean metric $d_e(x,y) = |x-y|$. Then $d'(x,y) = \max(d(x,y), |h(x) - h(y)|)$ (no need for $\sup$ with two real numbers) is a metric on $X$ as well and $(x,d')$ is equivalent to $(X,d)$. 
Proof: being a metric is easy. Clearly values of $d'$ are reals $\ge 0$ and $d'(x,y) = 0$ implies $d(x,y) =  0$ and as $d$ is a metric: $x=y$. Symmetry is clear as
$d'(x,y) = \max(d(x,y), |h(x) -h(y)|) = \max(d(y,x), |h(y) - h(x)|) = d'(y,x)$ using that $d$ and $|\cdot|$ are symmetric. For $ x,y,z$ in $X$ we have $d(x,z) \le d(x,y) + d(y,z)$ and $|h(x) - h(z)| = |h(x) - h(y) + h(y) - h(z)|  \le |h(x) - h(y)| + |h(y) - h(z)|$ as $d$ and the usual $|\cdot|$ metric on the reals are both metric, so the same inequality holds for their maxima ($a \le b, c \le d \rightarrow \max(a,c) \le \max(b,d)$ for any $a,b,c,d$ in a linearly ordered set), and so $d'$ obeys the triangle inequality as well.
The equivalence follows, because $d \le d'$ which again ensures that we can take $s=r$ in 2. 
To see 1. we consider $x \in X$ and $r> 0$. Then continuity of $h$ gives us that there is some $\delta >0 $ such that for all $y \in X$ we have that $d(x,y) < \delta$ implies $|h(x) - h(y)| < r$. Now define $s = \min(r, \delta)$. Then If $y \in B_{d}(x,s)$ we know that $d(x,y) < s$. So in particular $d(x,y) < r$ and $d(x,y) < \delta$, so $|h(x) - h(y)| < r$ as well. It follows that $d'(x,y) = \max(d(x,y), |h(x) -h(y)|) < r$ as well (whichever of the two numbers is the actual maximum, it's always $< r$). So $y \in B_{d'}(x,r)$ as required: we have shown $B_d(x,s) \subseteq B_{d'}(x,r)$.
So the two metrics have the same open sets; they are equivalent. All of this applies to your case, where $h(x) = \frac{1}{f(x)}$ as well.
