If we apply a homeomorphism on these objects, do they keep their properties. Let $f: (X, d_{X})\rightarrow (Y, d_{Y})$. And let:
a)$(u_n)$ is a convergent sequence of $(X, d_X)$
b)$(u_n)$ is a bounded sequence of $(X,d_X)$
c)$(u_n)$ is a Cauchy sequence of $(X,d_X)$
For $a)$, I didn't have any problems, as by the continuity of $f$ we have $\underset{n \rightarrow \infty}{\lim u_n}=l \iff \underset{n \rightarrow \infty}{\lim f(u_n)}=f(l) $. So the answer is yes, the property is kept.
Now, for $b)$, I am rather unsure. I have a hard time finding a counter example, and don't know how to prove that the property are kept.
Finally, for $c)$, I think it will depend on whether $X,Y$ are complete. If they are both complete, then I can use the fact that then $u_n$ converges to some $l \in X$ and : $$d_Y(f(u_n), f(u_m)) \leq d_Y(f(u_n), f(l)) + d_Y(f(l),f(u_m))$$ Would it be true if for example one is complete and another isn't? Or if both aren't complete?
EDIT 1: So based on the hint given my Max, if I put $(X, d_X)= (\mathbb{R}, |\cdot |)$ and then take $(Y, d_Y)(\mathbb{R}, \frac{|\cdot|}{1 + |\cdot|})$, then I am not sure what the homeomorphism would be equal to? Would $f(x)=x$ be enough? The sequence ould definitely be bounded on $(\mathbb{R}, \frac{|\cdot|}{1 + |\cdot|} )$, and I think a Cauchy sequence on the first metric would be a Cauchy sequence on the second metric space.
Consider $X = Y = (0,\infty)$, both endowed with the absolute value as metric. Let $f: X \to Y$ be given by $f(x)=1/x$. It's easy to see that $f$ is a homoemorphism.
Then consider the sequence $\{x_n\}_{n=1}^\infty \subset x$ given by $x_n = 1/n$. Then $f(x_n) = n$. This example handles the latter two of your three questions.