Proving the continuity of functions from metrics Context:
I'm studying mathematics at university, and am having trouble with some of the continuity questions. The following is a question from a previous assignment that I was unable to complete. The original question asked to first prove that $\bar\rho$ is a metric, but I've completed this part and adjusted the question accordingly. I understand the concept of the $\epsilon$-$\delta$ proofs, but I struggle when it comes to actually proving anything, and I'm never sure where to start. Any assistance on how to complete this would be appreciated. 
Question:
Let $(X,\rho)$ and $(X,\bar\rho)$ be metric spaces, where
$$
\bar\rho: X\times X\to\Bbb R_0^+\qquad(x,y)\mapsto\frac{\rho(x,y)}{1+\rho(x,y)}.
$$
Let $(Y,\sigma)$ be any metric space. Take functions $f: X\to Y$ and $g: Y\to X$.
Prove that:
(a) $f$ is continuous with respect to $\bar\rho$ if and only if it is continuous with respect to $\rho$; and,
(b) $g$ is continuous with respect to $\bar\rho$ if and only if it is continuous with respect to $\rho$.
 A: I proof the a) part, with this help b) can still be done as an exercise.
Assume $f$ is continuous in $x \in X$ with respect to $\bar{\rho}$.
Take $\epsilon>0$ we have to show: $$\exists \delta>0: \rho(x,y)<\delta \Rightarrow \sigma(f(x),f(y))< \epsilon$$
We know: $$\exists \eta>0: \bar{\rho}(x,y)<\eta \Rightarrow \sigma(f(x),f(y))<\epsilon$$
And we have:
$$\rho(x,y)>0 \Rightarrow \rho(x,y)+1>1 \Rightarrow \frac{1}{1+\rho(x,y)}<1 \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}<\rho(x,y) $$ $$\Rightarrow \bar{\rho}(x,y) < \rho(x,y) $$
So if we take $\delta = \eta$ we have:
if $\rho(x,y)<\delta$ then $\bar{\rho}(x,y)<\delta=\eta$
So by what we know it follows that: $\sigma(f(x),f(y))<\epsilon$
So $f$ is continuous with respect to $\rho$

Assume $f$ is continuous in $x \in X$ with respect to $\rho$. Take $\epsilon>0$ 
So: $$\exists \eta>0: \rho(x,y)<\eta \Rightarrow \sigma(f(x),f(y))<\epsilon \: (1)$$ (By continuity of $f$ with respect to $\rho$).
Denote: $\bar{\rho}= \bar{\rho}(x,y)$ and $\rho = \rho(x,y)$. We have $(\frac{\bar{\rho}}{1-\bar{\rho}} = \rho$):
$$2\bar{\rho}<\rho \Leftrightarrow 2 \bar{\rho}<\frac{\bar{\rho}}{1-\bar{\rho}} \Leftrightarrow 2(1-\bar{\rho})<1 \Leftrightarrow -2\bar{\rho}< -1 \Leftrightarrow \bar{\rho}>1/2 $$
Now take $\delta = \frac{1}{2}min\{ 1,\eta \}>0$. Let $\bar{\rho}(x,y)<\delta$ we then have: $\bar{\rho}(x,y)<\delta\leq\frac{1}{2}$ By the previous inequalities we then see:
$$2 \bar{\rho}(x,y) \geq \rho(x,y)$$ 
Now: $$\rho(x,y)\leq 2 \bar{\rho}(x,y)<2 \delta\leq 2 \frac{\eta}{2} = \eta$$
So by $(1)$ we see: $$\sigma(f(x),f(y))<\epsilon$$
This concludes the proof because:
$$\bar{\rho}(x,y)<\delta \Rightarrow \rho(x,y) < \eta \Rightarrow \sigma(f(x),f(y))$$

I have used of this:
 $$\bar{\rho}(x,y) =\frac{\rho(x,y)}{1+\rho(x,y)}$$
$$\Leftrightarrow \rho(x,y) =\frac{\bar{\rho}(x,y)}{1-\bar{\rho}(x,y)}$$
