There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$. Prove that, if $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

Proposition: If $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.
A (hypothesis): $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$.
B (conclusion): There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

My Work
B1: $(a + bi)(c + di) = 1$
$\therefore a + bi \not = 0$
B2: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$
A1: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$
$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$.
Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).
Now, below, I use the direct uniqueness method to show that the object is unique.
A2: There is a complex number $x + yi$ such that $x + yi = \dfrac{1}{a + bi}$.
$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$
A3: $(a+bi)(x + yi) = (a + bi)(c + di)$ where $a + bi \not = 0$
$\implies x + yi = c + di$ 
$Q.E.D.$

EDIT: My Work #2
A1: $(a + bi)(c + di) = 1$
$\implies ac + adi + bic + bdi^2 = 1$
$\implies ac + adi + bic - bd = 1$
$\implies (ac - bd) + i(ad + bc) = 1$ is a complex number, where $(ac - bd)$ is the real part and $i(ad + bc)$ is the imaginary part.
A2: Let $ac - bd = 1$ and $ad + bc = 0$
$\therefore 1 + 0 = 1$
Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).
Now, below, I use the direct uniqueness method to show that the object is unique.
A3: Let $(a + bi)(x + yi) = 1$
A4: $(a + bi)(c + di) = (a + bi)(x + yi)$ where ($a \lor b) \not = 0$.
$\implies c + di = x + yi$ 
$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback. If there are any errors, please explain why and what the correct procedure is.
 A: The aim of this exercise is to prove that the thing that you denote by $\frac1{a+bi}$ uniquely exists. Using this notation a priori implicitly assumes the statement we are about to prove.
Hint: Instead, as commented, try to explicitly find $c$ and $d$. For this, observe that $(a+bi)(a-bi)$ is real, and division is already known among reals.
For uniqueness of such $c+di$, we can take a general argument: 
Suppose $w_1$ and $w_2$ are both inverses of $z$, i.e. $w_1z=zw_2=1$.
Then we have
$$w_1 = w_1\cdot 1 = w_1zw_2=1\cdot w_2=w_2$$
A: I will start by trying to answer this explicit part of your question.

I would greatly appreciate it if people could please take the time to
  review my proof for correctness and provide feedback. If there are any
  errors, please explain why and what the correct procedure is.

In your first attempt you begin with the statement
$$
(a+bi)(c+di) = 1 .
$$
Since all you know from the hypothesis is the number $a+bi$ that beginning statement makes no sense. There is no $c+di$ in the hypothesis. No correct proof could start that way.
About your A1 you say

Therefore, in A1, I have constructed the object specified in the
  conclusion (c+di).

but that's not true. You have just written "$1/(a+bi) = x+yi$". To construct it you have to tell us the values of $x$ and $y$ in terms of $a$ and $b$. 
Where should you start? I suggest thinking about an example. Suppose you begin with $a+bi = 3+4i$. Then what you must do is find some (as yet unknown) complex number $c+di$ so that
$$
(3+4i)(c+di) = 1 .
$$
Well the second line in your A1 "proof" says you know what to do next:
$$
(3+4i)(c+di) = (3c -4d) + (3d + 4c)i = 1 .
$$
That tells you
$$
\begin{align}
3c - 4d & = 1 \\
3d + 4c & = 0
\end{align} 
$$ 
Now you can solve those two simultaneous equations in two unknowns and show that they have a unique solution.
Once you understand the example $3+4i$, you can do the same thing algebraically for the general complex number $a+bi$.
Hint. Your answer will have a square root in it. In the example that square root turned out to be an integer.
Note. This is a pretty longwinded way to go about the proof. There are cleverer shorter ways. But this is how you could approach the problem without knowing about the clever stuff.
Edit to answer the OP's last question.
There are several ways to show that the inverse $c+di$ is unique. One is to observe that the equations above you solved to find it have a unique solution. Another is to note that for any three complex numbers $\alpha$, $\beta$ and $\gamma$ with $\alpha \ne 0$,  if 
$$
\alpha \beta = \alpha \gamma
$$
then
$$
\alpha (\beta - \gamma) = 0
$$
so
$$
 \beta - \gamma  = 0
$$
(not because you "divide by $\alpha$" - because you check the arithmetic)
hence
$$
 \beta = \gamma.
$$
A: It seem that both proofs are strange or incomplete.
$\dfrac1{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i$
This is a unique complex number.
