Open cover; why is the following open? I am asked to look at the following

For $n=0,1,2...$ define $U_n \subseteq [0,1]$ by $U_0=[0,\frac{1}{2})$ and $U_n=(2^{-n},1]$ for $n \geq 1$. 

So, along the unit interval, we have a cover from zero to half and other sets $U_n$ covering $1/2,1/4...$ to $1$. Here's a question I cannot answer

Then $(U_n)_{n \geq 0}$ is an open cover of $[0,1]$. Why is this open?

No idea. Rather, aren't they..."clopen"? Clearly, each $U_n$ has one side "open" which is "(" or ")" but the other end is always "closed" "[ or ]."
Thus aren't each of $U_n$ ..clopen? So neither opened nor closed. So I am forced to challenge the statement above; they aren't open.
I checked the definition of open in this context but indeed,

$(U_i)_{i \in I}$ is open if $U_i$ is open for each $i \in I$.

Well, immediately, for $i = 1$, $U_1=(\frac{1}{2},1]$ which is not open(nor closed) so I've found a counterexample.
Please tell me why I am wrong! 

They are open in the subspace topology of $[0,1]$
For example, in your example $U_1 = (\dfrac{1}{2}, 1]$ is open, because it is an intersection of an open set in $\mathbb{R}$ say $(\dfrac{1}{2}, \dfrac{3}{2})$ with $[0,1]$
