I am trying to calculate the Hausdorff measure of the unit interval. Here's my attempt:
Fix $\epsilon>0$.
consider the open balls $B(x,\epsilon)$ with $x\in[0,1]$. How many can cover the unit interval? at least $\frac{1}{2\epsilon}$. So if we use $\lfloor{}\frac{1}{2\epsilon}\rfloor{}+1$ $\epsilon$-balls we will definitely cover the interval. So we have:
$$H^\delta([0,1])\le{}H^\delta_\epsilon([0,1])\le{}\sum_{i=1}^{\lfloor{}\frac{1}{2\epsilon}\rfloor{}+1}{(2\times\epsilon)^\delta}=(\lfloor{}\frac{1}{2\epsilon}\rfloor{}+1)(2\epsilon)^\delta\le{}(\frac{1}{2\epsilon}+1)(2\epsilon)^\delta$$
But here, I don't think this works as taking the limit $\epsilon\to0$ blows up the upper bound Is there any way to lower bound this to show that it is 1?
(answer for the original question.)
Take $\delta=1$, then the limit $\epsilon \to 0$. Your calculation shows $$ H^1([0,1]) \le 1 $$ Proof of the lower bound will also be needed: $$ H^1([0,1]) \ge 1 $$ But this is much harder. It can be done using properties of the Lebesgue measure. It is often true in estimation of Hausdorff measure. The upper bound it easy, the lower bound is hard.
Now a comment for the edited question. With $\delta < 1$ you should get $H^\delta([0,1]) = \infty$. Your calculation shows $$ H^{\delta}([0,1]) \le +\infty $$ which is not very useful. You will need to prove a lower bound, $$ H^{\delta}([0,1]) > 0 $$ As I remarked before, the lower bound is the harder one to prove.
It is easy to show $H^\delta(\mathbb Q \cap [0,1]) = 0$, so then you will know that the measures of the set and its closure are different.