3-SAT with 3 variable occurences
Your second step isn't sound.

Take any unsatisfiable $3$-SAT formula (without restriction on the number of variable appearances) and perform the standard reduction to a formula where each variable occurs at most three times.  That is, for each variable&nbsp;$X$ that occurs $k&gt;3$&nbsp;times, replaces the occurrences respectively with $X_1, \dots, X_k$ and add the new clauses

$$(\overline{X_1}\lor X_2) \land(\overline{X_2}\lor X_3) \land \dots \land (\overline{X_{k-1}} \lor X_k) \land (\overline{X_k}\lor X_1)$$

to ensure that $X_1, \dots, X_k$ have the same value in any satisfying assignment.

Now, we have the clause $\overline{X_1}\lor X_2$ but not $X_1\lor\overline{X_2}$ so we can use your second step, taking $A=\overline{X_1}$ and $B=X_2$. So we must delete the first clause and replace any instance of $X_1$ with&nbsp;$X_2$ and any instance of $\overline{X_2}$ with&nbsp;$\overline{X_1}$.  This replaces the clauses above with 

$$(\overline{X_1}\lor X_3) \land (\overline{X_3}\lor X_4) \land \dots \land (\overline{X_{k-1}} \lor X_k) \land (\overline{X_k}\lor X_2)\,,$$
i.e.,
$$(X_1\rightarrow X_3)\land (X_3\rightarrow X_4) \land \dots \land (X_{k-1}\rightarrow X_k)\land (X_k\rightarrow X_2)\,.$$

This no longer requires all the the $X_i$ variables to have the same value. There are formulas that will become satisfiable when you make this transformation.

But, further, we can now apply the transformation again to the clause $\overline{X_1}\lor X_3$, resulting in
$$(\overline{X_1}\lor X_4) \land (\overline{X_4}\lor X_5)\land\dots\land (\overline{X_{k-1}} \lor X_k) \land (\overline{X_k}\lor X_2)\,.$$
Note that the variable $X_3$ has completely disappeared and can take any value. Iterating, we see that all the clauses relating $X_1, \dots, X_k$ disappear!

What have we done? We've taken a formula that mentioned&nbsp;$X$ more than three times and replaced these occurrences with successive new variables $X_1, \dots, X_k$, plus the constraint that all these new variables must take the same value.  We then deleted all the constraints, so $X_1, \dots, X_k$ can take any value at all, independently of each other and, by the way, each of the variables $X_1, \dots, X_k$ now only occurs once.  So, if we started with an unsatisfiable formula in which every variable occurred more than three times, we end up with a formula where every variable occurs exactly once.  That formula is always satisfiable.