Interplay of Hausdorff metric and Lebesgue measure

Counterexample. In one dimension, let $K_k=\{0,1/k,2/k,3/k,\dots,1\}$. Each $K_k$ has zero measure. The limit of $K_k$ in the Hausdorff metric is the interval $[0,1]$, of measure $1$.

You can also attach the same interval to the sets $K_k$ to make their measure positive.

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The above counterexample could be ruled out by requiring $K_k = \overline{\operatorname{int} K_k}$. But then there's a different one: $K_k = \bigcup_{j=1}^k [(2j-1)/2k,j/k] $ where each set has measure $1/2$ but the limit is again $[0,1]$.

I don't think you can benefit from Hausdorff-metric convergence here at all. I suggest looking at sufficient conditions for $L^1$ convergence, so that you can get $\chi_{K_k}\to \chi_K$ in $L^1$.

As pointed out, the answer is negative. However, Lebesgue measure is upper semi-continuous with respect to the Hausdorff metric. That is, the measure can only increase in the limit, not decrease.

Indeed, suppose $X_n\to X$ in the Hausdorff metric, and $C:= \lambda(X)<\infty$. Let $U\supset X$ be an open set of measure at most $C+\varepsilon$. Then, for sufficiently large $n$, $X_n\subset U$, and hence $\lambda(X_n)\leq C+\varepsilon$. So indeed $$ \limsup_{n\to\infty} \lambda(X_n) \leq \lambda(X),$$ as claimed.

In particular, if $X$ is a null-set, then $X_n$ is also a null-set for sufficiently large $n$. As pointed out in the accepted answer, this is the only case where the answer to your question is positive, since any compact set is the Hausdorff limit of a sequence of finite sets.