Find area of unit square using outer Hausdorff-measure We put $$\eta_{\delta}(E) = \inf\left\{ \sum_{i \in \mathbb{N}} \text{diam }U_i : E \subset \bigcup_{i\in \mathbb{N}} U_i \text{, and diam }U_i\in(0,\delta] \right\}$$
and for $E\subset \mathbb{R}^2$ we define the outer Hausdorff-measure on $\mathbb{R}^2$ as $\eta = \lim_{\delta \to 0} \eta_{\delta} = \sup_{\delta>0} \eta_{\delta}$.
Find the value of $\eta( \{ (0,0) \} ), \eta( [0,1] \times \{0\})$ and $\eta([0,1]^2).$
So far I've proved that $\eta( \{ (0,0) \} )=0$ by choosing the covering $(U_i)_{i \in \mathbb{N}}$ where $U_i = [0, 2^{-i} \delta) \times \{0\}$. Then 
$$\begin{eqnarray}
\eta( \{ (0,0) \} ) &=& \lim_{\delta \to 0} \eta_{\delta}( \{ (0,0) \} )
\\&\leq& \lim_{\delta \to 0} \sum_{i \in \mathbb{N}} 2^{-i} \delta
\\&=& \lim_{\delta \to 0} \delta
\\&=& 0.
\end{eqnarray}$$
But so far I have failed for the other two. I'm very grateful for any type of hint.
 A: Consider $E:=[0,1]\times\{0\}$. We claim that $\eta(E)=1$. 
For $\delta>0$ let $N=\lfloor\frac1\delta\rfloor+1$, so that $\frac 1N<\delta\le\frac1{N-1}$.
For $1\le i\le N$ let $U_i=B\left((\frac {2i-1}{2N},0),\frac12\delta\right)$. Then $\operatorname{diam}(U_i)=\delta$ and $\sum \operatorname{diam}(U_i)=N\delta<1+\delta$. This shows $\eta_\delta(E)\le 1+\delta$ and $\eta(E)\le 1$.
On the other hand, let $\{U_i\}_{i\in\mathbb N}$ be any open cover of $E$
and assume $\sum\operatorname{diam}(U_i)<1$.
By compactness of $E$, a finite subcover $\{U_i\}_{1\le i\le N}$ suffices and wlog. all $U_i\cap E$ are nonempty.
Recall that $U\cap V\ne \emptyset$ implies $\operatorname{diam}(U\cup V)\le \operatorname{diam}(U)+\operatorname{diam}(V)$.
Therefore, we may repeatedly replace two overlapping elements of our finite cover with their union, until after finitely many steps we arrive at a finite open pairwise disjoint cover of $E$ that still has diameter sum $<1$. Since  $E$ is connected, the only possibility is that the remainig cover consists of a dingle open set. But then its diameter must be at least $1$ as it contains $(0,0)$ and $(1,0)$. Hence $\sum\operatorname{diam}(U_i)\ge 1$ for any open cover of $E$, so that we ultimattely find $\eta(E)=1$.

Now consider $E=[0,1]\times[0,1]$.
For $n\in\mathbb N$ and $\delta<\frac1n$, observe that in an open cover $\{U_i\}$ of $[0,1]\times[0,1]$, any $U_i$ intersects at most one of the horizontal lines $[0,1]\times\{\frac kn\}$, $0\le k\le n$, because its diameter is too small to touch two such lines. Using the previous result, we find
$$\sum_{i\in\mathbb N}\operatorname{diam}(U_i)\ge \sum_{k=0}^n\sum_{i\in \mathbb N\atop U_i\cap [0,1]\times\{\frac kn\}\ne\emptyset}\operatorname{diam}(U_i) \ge \sum_{k=0}^n1=n+1.$$
As we let $\delta\to0$ this shows $\eta([0,1]\times[0,1])=\infty$.
