Show that $A$ is closed in $X$ and $f(A)$ is not closed in $Y$.

Question

Let $X=[0,1)$ with the metric $d(x,y)=|x-y|$, and $Y=\mathbb{R}^2$ with the Euclidean metric. Define the mapping $f:X\rightarrow{Y}$ by

$f(t)=(\cos(2 \pi t + \frac{\pi}{2}), \sin(2 \pi t + \frac{\pi}{2})) \forall t\in{X}$.

Let $A=[1/2,1)$. Show that $A$ is closed in $X$ and $f(A)$ is not closed in $Y$.

Obviously $A$ is closed in $X$, but I don't know how to show that mathematically. Should I somehow show the complement of A is open?

For $f(A)$ not being closed in $Y$, I plugged in the values $\frac{1}{2}$ and $1$ into $f(A)$ and found the range to be $(-1,1)$.

Not sure if I'm on the right track... any hints and/or direction would be greatly appreciated!