Hausdorff dimension of Cantor set

I can elaborate abit why it is sufficient to show it for open ones. It might be that this is not exactly what the author was going after. If you wish, I may also continue this answer into a full proof which shows that $H^{\alpha}(C)\geq \frac{1}{2}$.

Choose for starters a $\delta$-cover $\{E_{j}\}_{j=1}^{\infty}$ of $C$ with $\sum_{j=1}^{\infty}\mathrm{diam}(E_{j})^{\alpha}\leq H^{\alpha}(C)+\delta$. Then for each $j$ we may choose a closed interval $I_{j}$ with $E_{j}\subset \mathrm{int}I_{j}$ and $\mathrm{diam}(I_{j})<(1+\alpha)\mathrm{diam}(E_{j})$. Hence $\{\mathrm{int}I_{j}\}_{j=1}^{\infty}$ is an open cover of $C$, and in particular \begin{align*} H^{\alpha}(C)+\delta\geq \sum_{j=1}^{\infty}\mathrm{diam}(E_{j})^{\alpha}\geq (1+\delta)^{-\alpha}\sum_{j=1}^{\infty}\mathrm{diam}(I_{j})^{\alpha}. \end{align*} Thus to establish a lower bound for $H^{\alpha}(C)$ it suffices to establish a lower bound for $\sum_{j=1}^{\infty}\mathrm{diam}(I_{j})^{\alpha}$. (You may consider $\mathrm{int}(I_{j})$'s instead, which are open, as their diameter is the same as $I_{j}$'s.)

It is a fact in general (i.e. true in any metric space and Hausdorff measures of any dimension) that you can assume your covering sets to be open or closed. See Federer. For closed version, it is easier: diameter of $\bar{S}$ and diameter of $S$ are equal, so, if a collection of $S$'s cover your set and each has diam less than $\delta$, then you can instead consider the collection of $\bar{S}$'s, which again have diam bounded by $\delta$ ans still cover your set.

For open version, you need some sacrifice! At a cost of arbitrarily small $\sigma$, you can enlarge every set of diam less than $\delta$ to an open one with diam less than $(1+\sigma)\delta$. The latter can be used to estimate $\mathcal{H}^s_{(1+\sigma)\delta}$ within $(1+\sigma)^s$ accuracy of $\mathcal{H}^s_{\delta}$. Since for Hausdorff measure $\mathcal{H}^s$, you will send $\delta \to 0$, and $(1+\sigma)\delta$ will as well, your sacrifices will not affect the ultimate measure. (However, as expected, for one fixed $\delta$, $\mathcal{H}^s_\delta$ can be different if you only allow open coverings versus all coverings.)