Let $c \in \mathbb{R}$ and $X=(c, \infty)$, $Y=(-\infty,c).$ Show that $X$ is homeomorphic to $Y$.
To show that $X$ is homeomorphic to $Y$ I would need to find a bicontinuous function $f :X \to Y$?
Define $f(x)= 2c-x$. If $x \in X$, then $f(x) < c$ thus $f(x) \in Y$. Define $g(y) =2c-y$ if $y \in Y$, then $g(y) > c$ so $g(y) \in X$. So I have that $f : X \to Y$ and $g : Y \to X$ also $$g \circ f=g(f(x))=g(2c-x)=2c-(2c-x)=x \\ f\circ g=f(g(y)) = f(2c-y)=2c-(2c-y)=y$$ which implies that $g = f^{-1}$. To show that this is indeed bicontinuous I have that $$|f(x)-f(y)|=|(2c-x)-(2c-y)|=|x-y|, x,y\in X$$ which would imply that $f$ is bi-lipschitz and thus homeomorphic. Is this a valid way to show this?
Sure, this proof is fine. Instead of Lipschitz you could also use the order-reversing property (and the fact that these sets have the order topology).