Algebraic treatment of equilibrium
There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid $\ce{HA}$ results in equal amounts of hydronium and of conjugate base $\ce{A-}$. Unless a salt of the conjugate base is explicitly added then the acid is the only source of $\ce{A-}$. In absence of salt the total hydronium concentration in solution is therefore 

$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$ 

Note that this equation reflects the charge balance condition.

Solving for $\ce{[A-]}$

$$\ce{[A-]} = \ce{[H+]}-\ce{[OH-]}\tag{1}$$ 

We also know that the original amount of acid equals the sum of undissociated acid and conjugate base:  

$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$ 

Note that this equation reflects the mass balance condition.

Solving for $\ce{[HA]}$

$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$ 

Substituting Eq. (1) into Eq. (2) we obtain  

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$

The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution ($C_\ce{A-}=0$). If a completely dissociating salt $\ce{MA}$ is added at concentration $C_\ce{A-}$ then it is necessary to modify the mass balance expression:

$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$ 

and the charge balance expression: 

$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$ 

or since $\ce{[M+]}=C_\ce{A-}$

$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$ 

Solving for $\ce{[A-]}$

$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$ 

Combining the two balance expressions leads once again to Eq. (3):

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$

By the way, the quote you cite should read:

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

It might be that "typo" ("3" written instead of "2") that led to the confusion.