Let $f$ be a function from the one-point compactification of $\mathbb{R}$ to the circle defined as follows
$t\mapsto (x,y) = \begin{cases} \phantom{\lim\limits_{t\to\infty}} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t\ne\infty, \\[10pt] \lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t=\infty. \end{cases}$
I'm having trouble proving continuity and that it's open. Can you help me?
N/A = I tried many things, but none had a sense or took to somewhere. I actually need help...
Continuity is a local concept, so the only thing to check here is that $f$ is continuous at $\infty$ (continuity at all other points is obvious). Note that $f(\infty) = (-1,0)$. By definition of the topology in the one-point compactification, it suffices to show that if $t \to \pm \infty$, then $f(t) \to (1,0)$, but this holds pretty much from the way $f$ is defined.
Once you have shown that $f$ is continuous, verify that it is bijective. Then you are done, because a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism, as continuity of the inverse follows from $f$ being closed.
Alternatively, you can exhibit a formula for the inverse, which will be manifestly continuous. For any point $p$ in the circle not equal to $(-1,0)$, consider the ray starting at $(-1,0)$ and passing through $p$. This ray will cross the $y$-axis at a single point -- the $y$-coordinate of such point is $f^{-1}(p)$, and then you set $f^{-1}(-1,0) = \infty$.