In the proof of the Hausdorff dimension of the middle third cantor set I cannot understand why we need the following underlined statement. I cannot understand why we need only consider closed intervals? Moreover, I am not sure in the construction why we would include/exclude the end points of the intervals.
In this proof $|U|$ is the diameter of $U$. It isn't hard to see that for any $\epsilon > 0$ there and any set $U$ with finite diameter there is an open interval $V$ with the property that $U \subset V$ and $|V| = (1 + \epsilon)|U|$. To see this, observe that if $U$ has finite diameter then $U$ is bounded, $U \subset [\inf U, \sup U]$, and $|U| = \sup U - \inf U$. Define $V = (\inf U - \delta,\sup U + \delta)$ where $\delta > 0$ is chosen so that $|V| = |U| + 2\delta = (1 + \epsilon)|U|$.
Let $\{U_i\}$ be a cover of $F$ by sets with finite diameter. Let $\epsilon > 0$ be given and select the $\{V_i\}$ accordingly. Since $F$ is compact there exists an index $I$ so that $\{V_i\}_{i=1}^I$ is a cover of $F$. If in this instance you can verify that $\sum_{i=1}^I |V_i|^s \ge 3^{-s}$ you will obtain $$ (1+\epsilon)\sum_i |U_i|^s = \sum_i |V_i|^s \ge \sum_i |V_i|^s \ge 3^{-s} $$ for the arbitrary cover. Then let $\epsilon \to 0^+$.