Solution:

Let $A$ and $B$ be the subsets selected by Kyle and Albert, respectively. We first find the probability that two subsets $A$ and $B$ are disjoint. For each $k \in \{0,1, \ldots, 4\}$, we choose an arbitrary subset $A$ with $k$ elements. In order for $A$ and $B$ to be disjoint, $B$ must be a subset of the complement $\{1,2,3,4\} \backslash A$ with $4-k$ elements. Thus, for each $k \in \{0,1, \ldots, 4\}$, there are $\binom{4}{k}$ subsets with $k$ elements and fixing one of such subsets (say $A$), there are $2^{4-k}$ choices for $B$ (note that $2^{4-k}$ is the number of subsets of $\{1,2,3,4\} \backslash A$). We see that the number of ordered pairs $(A, B)$ of subsets with $A \cap B = \varnothing$ is
$$\sum_{k=0}^{4} \binom{4}{k} 2^{4-k} = 16 \sum_{k=0}^{4} \binom{4}{k} \left(\frac{1}{2}\right)^k = 16\left(1+\frac{1}{2}\right)^4 = 81$$
As there are $2^4 = 16$ subsets of $\{1,2,3,4\}$, there are $16^2 = 256$ possible ordered pairs of subsets. Hence, the probability that two subsets $A$ and $B$ are disjoint is $\frac{81}{256}$ and the probability that $A$ and $B$ have at least one element in common is $1 - \frac{81}{256} = \frac{175}{256}$.