Phicar’s answer gives you a nice, short calculation if you know about Stirling numbers of the second kind. If not, you can still organize your argument a bit more efficiently.
Suppose that the set $A\cup B$ has $n$ elements; clearly $n$ must be $2,3,4$, or $5$. For each of these four possible values of $n$ we can argue as follows.
There are $\binom5n$ ways to choose the set $A\cup B$. $A$ can be any non-empty proper subset of $A\cup B$. $A\cup B$ has $2^n$ subsets, but one is empty and one is all of $A\cup B$, so there are only $2^n-2$ choices available for $A$. Thus, there are $\binom5n(2^n-2)$ ordered pairs $\langle A,B\rangle$ with $|A\cup B|=n$.
The answer, therefore, is
$$\sum_{n=2}^5\binom5n(2^n-2)=\sum_{n=2}^5\binom5n2^n-2\sum_{n=2}^5\binom5n\;.$$
Now notice that
$$\sum_{n=2}^5\binom5n=\sum_{n=0}^5\binom5n-\binom51-\binom50=2^5-5-1=26\;,$$
and
$$\begin{align*}
\sum_{n=2}^5\binom5n2^n&=\sum_{n=0}^5\binom5n2^n-\binom512-\binom50\\\\
&=\sum_{n=0}^5\binom5n2^n1^{5-n}-10-1\\\\
&=(2+1)^5-11\\\\
&=232\;,
\end{align*}$$
so the final answer is $232-52=180$.
This calculation suggests another elementary way to perform the calculation. If we temporarily allow $A$ and $B$ to be empty, we are in effect counting the ways to split $X$ into $3$ pieces, any of which may be empty. For each of the $5$ elements of $X$ we can put that element into $A$, into $B$, or into $X\setminus(A\cup B)$. This is a $3$-way choice made $5$ times, so there are $3^5=243$ ways to make it. However, some of these splits leave $A$ or $B$ or both empty. We can use an inclusion-exclusion argument to take care of them.
How many of these splits leave $A$ empty? They are the splits that put every element into $B$ or $X\setminus(A\cup B)$, so there are $2^5$ of them. There are also $2^5$ splits leaving $B$ empty, so we have to subtract $2\cdot2^5=64$. However, that subtracts the one split with $A=B=\varnothing$ twice, so we have to add it back in. The final result is
$$243-64+1=180\;.$$
For the first part of the problem, we start with the observation that any unfriendly subset of $[n+2]$ either contains $n+2$ or does not contain $n+2$.
We count the number of unfriendly subsets in each case separately.
We first count all unfriendly subsets of $[n+2]$ that contain $n+2$. We do this by establishing a bijection between unfriendly subsets of $[n]$ and those that we just described. Note that by adding $n+2$ to each of the unfriendly subsets of $[n]$, we get unfriendly subsets of $[n+2]$. Conversely, every unfriendly subset of $[n+2]$ certainly does not contain $n+1$ and by removing $n+2$ from each of them, we get unfriendly subsets of $[n]$. This means that the number of unfriendly subsets of $[n+2]$ that contain $n+2$ is equal to the number of unfriendly subsets of $[n]$ (which is equal to $U(n)$, of course).
Now we count unfriendly subsets of $[n+2]$ that do not contain $n+2$. It is clear that each of the sets falling into this category is also an unfriendly subset of $[n+1]$. Therefore, the number of unfriendly subsets of $[n+2]$ that do not contain $n+2$ is equal to the number of unfriendly subsets of $[n+1]$ (which is equal to $U(n+1)$).
Therefore, we arrive at $U(n+2)=U(n+1)+U(n)$ for each $n\in\mathbb{N}$.
Now, by a similar argument, we have
$$U(n,k)=U(n-1,k)+U(n-2,k-1)$$
for valid $n$ and $k$.
It is clear that $U(n,0)=1$ since the only unfriendly $0$-subset of $[n]$ is the null set.
Note that all singleton subsets of $[n]$ are unfriendly, and therefore $U(n,1)=n$.
The above recurrence relation for $U(n,k)$ can be rewritten as
$$U(n-2, k-1)=U(n,k)-U(n-1,k)$$
Summing over $n$, we obtain
$$\sum_{n=k+1}^m \: U(n-2,k-1) = \sum_{n=k+1}^m \:(U(n,k)-U(n-1,k))=U(m,k)-U(k,k)=U(m,k)$$
since $U(k,k)=0$ for all $k\geq 2$. Now, letting $k=2$ gives
$$U(m,2)=\sum_{n=3}^m \: U(n-2,1)=\sum_{n=3}^m \: (n-2) = 1+2+\dots+(m-2)=\binom{m-1}{2}$$
Letting $k=3$ gives
$$U(m,3)=\sum_{n=4}^m \: U(n-2,2)=\sum_{n=4}^m \: \binom{n-3}{2} =\binom{m-2}{3}$$
where we have used the infamous hockey-stick identity in the last step.
Now, it is easy to show (by induction on $k$) that
$$U(m,k)=\binom{m-k+1}{k}$$
Best Answer
First, find the number of unordered subsets of $[n]$ with no difference of 1, of size $k$. The number of these is ${n - k + 1 \choose k}$. If we have a subset $a_1, a_2, \ldots, a_k$ of $[n]$ in increasing order, with none of the differences 1, then $a_1, a_2 - 1, a_3 - 2, \cdots, a_k - (k-1)$ is a subset of $[n-k+1]$ in increasing order, and this transformation can be reversed. For example, consider the subsets of $[7]$ of size 3 with no difference 1; these are
$$135, 136, 137, 146, 147, 157, 246, 247, 257, 357$$
(where I write $135$ for $\{1, 3, 5\}$ for conciseness). We can subtract 1 from the second element and 2 from the third element of each of these to get
$$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$$
Both of these contain ${7 - 3 + 1 \choose 3} = 10$ elements.
Now, as you've already noticed, each unordered set of size $k$ gives rise to $k!$ ordered sets. For size $n$ we can have $k$ as large as $\lceil n/2 \rceil$. So we have
$$f(n) = \sum_{k=1}^{\lceil n/2 \rceil} {n-k+1 \choose k} k! = \sum_{k=1}^{\lceil n/2 \rceil} {(n-k+1)! \over (n-2k+1)!}$$
For numerical values, see https://oeis.org/A122852 (as has already been observed) and subtract one. For example,
\begin{align} f(7) &= \sum_{k=1}^4 {8-k \choose k} k! \\ &= {7 \choose 1} 1! + {6 \choose 2} 2! + {5 \choose 3} 3! + {4 \choose 4} 4! \\ &= 7 \times 1+ 15 \times 2 + 10 \times 6 + 1 \times 24 \\ &= 121 \end{align}