Your first interpretation is the correct one.
The topology on $K(X)$ induced by the Hausdorff metric has a description as a hit-and-miss topology. That is, for a nonempty open set $U \subseteq X$ put
$$
U^+ = \{C \in K(X) \mid C \cap U \neq \emptyset\}
$$
and
$$
U^- = \{C \in K(X) \mid C \subseteq U\}.
$$
The set $U^+$ contains the compact sets that meet $U$ and the set $U^-$ contains the compact sets that miss $X \setminus U$.
The sets $U^+$ and $U^-$ where $U$ runs through the non-empty open sets of $X$ form a subbasis for the so-called Vietoris topology on $K(X)$. The sets $[U; V_1,\dots,V_n] := U^- \cap V_{1}^+ \cap \cdots \cap V_{n}^+$ form a convenient basis for the Vietoris topology.
The point is that one can prove that the Hausdorff metric induces the Vietoris topology on $K(X)$.
It may be helpful to prove at some point that for a countable dense subset $D$ of $X$ the collection of non-empty finite subsets of $D$ forms a countable dense subset of $K(X)$.
A nice and detailed exposition of these ideas can be found e.g. in Srivastava, A course on Borel sets, section 2.6, see Spaces of Compact sets, pages 66ff.
Added: The second interpretation is too strong. It is not very hard to show that a fat Cantor set and the usual ternary Cantor set are homeomorphic, but not bi-Lipschitz homeomorphic.
The open ball $B(x_0,1)$ is defined as all the points whose distance from $x_0$ is less than one. Since any point other than $x_0$ has distance of exactly one from $x_0,$ (and one is not less than one) any point other than $x_0$ is not in $B(x_0,1).$ Thus $B(x_0,1) = \{x_0\}.$
Best Answer
I can show you a sequence $Y_n$ that converges to $Y=[-1,0]^2$ in the area metric, but not in the Hausdorff metric. Near the points $(\frac k n,0)$, $k=1,\dots, n-1$, consider a triangular spike of height $1$ and area $\frac{1}{n^2}$, for example the triangle with vertices $(\frac k n-\frac{1}{n^2},0), (\frac k n,1), (\frac k n+\frac{1}{n^2},0)$. Define $Y_n$ as the union of $Y$ and these triangles. Then the distance between $Y_n$ and $Y$ in the area metric tends to zero, but the distance in the Hausdorff metric does not.
For the positive result, try to consider two convex polygons and try to bound the difference of their areas in terms of the Hausdorff distance and vice versa. Even if the metrics are not equivalent, that may be an easier approach to showing the topologies are the same than considering open balls.
For example, if you have two sets whose Hausdorff distance is small, you have $A_1\subset B_\epsilon(A_2)$ and vice versa. This means that $A_1{\small\Delta} A_2\subset B_\epsilon(A_1)\setminus A_1\cup B_\epsilon(A_2)\setminus A_2$, and the area of that set is bounded by $C\epsilon$ times the perimeter of $A_1$.
On the other hand, if two convex polygons have positive Hausdorff distance, you need to show their area distance can't be arbitrarily small. I can visualise why that should be true, but I can't give you an even semi-formal proof at the moment.