Let $A$ be a subset of $\mathbb{R}^{2}$ and consider the projection on the x-axis as follows:
$$B=\pi(A)=\{ x \in \mathbb{R} : (x,y) \in A, y \in \mathbb{R} \}$$
I am asked to show that the 1-dimensional Hausdorff outer measure satisfies
$$\mathcal{H}^{1}(B) \leq \mathcal{H}^{1}(A)$$
I've tried taking a cover of $A$ made of 2-dimensional balls with radii less than $\delta$ and consider their projection on the x-axis as intervals, but I got stuck. I know that the collection of these intervals should be covering $A$, but I might be mistaken. Any help would be greatly appreciated.
First, note that that projections preserve containment. That is, if $U \subseteq V$, then $\pi(U) \subseteq \pi(V)$. To see this, suppose that $x \in \pi(U)$. Then there is some $y$ such that $(x,y) \in U$. But $U\subseteq V$, so $(x,y) \in V$, from which it follows that $x = \pi(x,y) \in \pi(V)$.
Next, observe that distances decrease under projection. This is not hard to see in the particular example of projection onto the $x$-axis: if $x$ and $y$ are points in $\mathbb{R}^2$, then $$ d(x,y) \ge d(\pi(x),\pi(y)). $$ To see this, observe that if $x = (x_1,x_2)$ and $y = (y_1,y_2)$, then $$ d(x,y)^2 = (x_1-y_1)^2 + (x_2-y_2)^2 \ge (x_1-x_2)^2 = (\pi(x)-\pi(y))^2 = d(\pi(x),\pi(y))^2. $$ This implies that projection onto the $x$-axis decreases distance, as claimed. This holds in greater generality, but it is not necessary to prove it here.
As the diameter of a set is, by definition, the supremal distance between pairs of points in that set, this implies that projections decrease the diameters of sets. Combining this with the fact that projections preserve containments, we can conclude that if $\{U_n\}$ is a $\delta$-cover of $A$, then $\{\pi(U_n)\}$ is a $\delta$-covering of $B = \pi(A)$. But then \begin{align} \mathcal{H}_\delta^1(A) &= \inf\left\{ \sum_{n=1}^{\infty} |U_n| : \text{$\{U_n\}$ is a $\delta$-cover of $A$} \right\} \\ &\ge \inf\left\{ \sum_{n=1}^{\infty} |\pi(U_n)| : \text{$\{U_n\}$ is a $\delta$-cover of $A$} \right\} \\ &\ge \inf\left\{ \sum_{n=1}^{\infty} |V_n| : \text{$\{V_n\}$ is a $\delta$-cover of $B$}\right\} \tag{$\ast$} \\ &= \mathcal{H}_\delta^1(B). \end{align} At ($\ast$), we use the fact that if $X \subseteq Y \subseteq \mathbb{R}$, then $\inf(X) \ge \inf(Y)$. Taking $\delta\to 0$, we obtain the desired result.