If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric

a) Separation and symmetry are clear. For the triangular inequality, there is a bit more work. Let $$f(t):=\frac{t}{1+t}\qquad f'(t)=\frac{1}{(1+t)^2}.$$ Since this function increases on $[0,+\infty)$, the triangular inequality of $d$ yieds $$ d_b(x,y)=f(d(x,y))\leq f(d(x,z)+d(z,y))=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)} $$ $$ \leq f(d(x,z))+f(d(z,y))=d_b(x,z)+d_b(z,y). $$

b) You need to show that a sequence converges to $x$ for $d$ if and only if it converges to $x$ for $d_B$.

Assume first that $d(x_n,x)\rightarrow 0$. Then $$ d_B(x_n,x)=\frac{d(x_n,x)}{1+d(x_n,x)}\leq d(x_n,x) $$ so $d_B(x_n,x)$ tends to $0$.

Now if $d_B(x_n,x)$ tends to $0$, $d(x_n,x)$ is bounded by some $M>0$. Indeed, assume for a contradiction that $d(x_n,x)$ is unbounded. So there exists a subsequence $d(x_{n_k},x)$ which tends to $\pm\infty$. Then $d_B(x_{n_k},x)$ must tend to $1$. Contradiction.

Now $$ \frac{d(x_n,x)}{1+M}\leq \frac{d(x_n,x)}{1+d(x_n,x)}=d_B(x_n,x). $$ So $d(x_n,x)$ tends to $0$.

c) Observe that $d_B$ is bounded while $|x-y|$ is unbounded. So such a minoration is impossible.

Separation, identity of indiscernibles and symmetry are immediate by the properties of $d$. For the triangular inequality, here is a simpler step-by-step proof without having to consider derivatives:

$$d_b(x,z) = \frac{d(x,z)}{1 + d(x,z)} = \frac{1 + d(x,z) - 1}{1 + d(x,z)} = 1 - \frac{1}{1 + d(x,z)}$$

$$\leq 1 - \frac{1}{1 + d(x,y) + d(y,z)} = \frac{1 + d(x,y) + d(y,z) - 1}{1 + d(x,y) + d(y,z)} $$

$$ = \frac{d(x,y)}{1 + d(x,y) + d(y,z)} + \frac{d(y,z)}{1 + d(x,y) + d(y,z)}$$

$$\leq \frac{d(x,y)}{1 + d(x,y)} + \frac{d(y,z)}{1 + d(y,z)} = d_b(x,y) + d_b(y,z) $$

where the first inequality is from the sub-additivity of $d$, and the second inequality is from the fact that reducing a positive denominator can only increase a positive quotient.