Diagonal Scaling of $ A \in {\mathbb{R}}^{2 \times 2} $ Positive Definite and Its Conditional Number Given a Positive Definite Matrix $ A \in {\mathbb{R}}^{2 \times 2} $ given by:
$$ \begin{bmatrix}
{A}_{11} & {A}_{12} \\ 
{A}_{12} & {A}_{22}
\end{bmatrix} $$
And a Matrix $ B $ Given by:
\begin{bmatrix}
\frac{1}{\sqrt{{A}_{11}}} & 0 \\ 
0 & \frac{1}{\sqrt{{A}_{22}}}
\end{bmatrix}
Now, defining the Diagonal Scaling of $ A $ given by $ C = B A B $.
One could see the main diagonal elements of $ C $ are all $ 1 $.
Actually $ C $ is given by:
$$ C = \begin{bmatrix}
1 & \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} \\ 
\frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} & 1
\end{bmatrix} $$
This scaling, intuitively, makes the Matrix closer to the identity Matrix (Smaller off diagonal values, 1 on the main diagonal) and hence improve the Condition Number.
Yet I couldn't prove it.
How could one prove $ \kappa \left( C \right) = \kappa \left( B A B \right) \leq \kappa \left( A \right) $?
 A: The condition number and positive definiteness is independent of scaling the matrix by a positive scalar and the result is obviously true for diagonal matrices, so we may assume that
$$
A:=\pmatrix{\alpha&1\\1&\beta}, \quad \alpha,\beta>0, \quad 1<\alpha\beta
$$
(if the off-diagonal entry is negative, apply a similarity transformation with the diagonal matrix $\mathrm{diag}(1,-1)$).
Then
$$
\kappa(A)
=\frac{\alpha+\beta+\sqrt{(\alpha-\beta)^2+4}}{\alpha+\beta-\sqrt{(\alpha-\beta)^2+4}}
=\frac{1+\sqrt{\frac{(\alpha-\beta)^2+4}{(\alpha+\beta)^2}}}{1-\sqrt{\frac{(\alpha-\beta)^2+4}{(\alpha+\beta)^2}}}
$$
For the scaled matrix, we have
$$
\kappa(C)=\frac{1+\frac{1}{\sqrt{\alpha\beta}}}{1-\frac{1}{\sqrt{\alpha\beta}}}.
$$
Since $1<\alpha\beta$, you can show that both expressions are of the form
$\frac{1+\sqrt{t}}{1-\sqrt{t}}$ where $t\in[0,1)$. Since this function is increasing on this interval, we just need to verify that 
$$
\frac{1}{\alpha\beta}\leq \frac{(\alpha-\beta)^2+4}{(\alpha+\beta)^2}
$$
(again, using the assumptions on $\alpha$ and $\beta$).

By the way, your intuition might mislead you in the general case. Such a diagonal scaling might fail even for SPD matrices of size at least $3$.
A: Since multiplying by a constant doesn't change the Condition Number one could make the choice of:
$$ A = \begin{bmatrix}
a & 1 \\ 
1 & b
\end{bmatrix} $$
Hence the matrix $ B $ becomes:
$$ B = \begin{bmatrix}
\frac{1}{\sqrt{a}} & 0 \\ 
0 & \frac{1}{\sqrt{b}}
\end{bmatrix} $$
And the Matrix $ C $ becomes:
$$ C = \begin{bmatrix}
1 & \frac{1}{\sqrt{ab}} \\ 
\frac{1}{\sqrt{ab}} & 1
\end{bmatrix} $$
Moreover, defining $ {\alpha}_{1} \geq {\alpha}_{2} $ are the Eigen Values of $ A $ and $ {\gamma}_{1} \geq {\gamma}_{2} $ are the Eigen Values of $ C $.
One should notice that $ \det \left( B \right) = \frac{1}{\sqrt{ab}} < 1 $ by the definition of $ C \succ 0 $ (By Quadratic Form of PD Matrix) or $ A \succ 0 $ which implies $ a b > 1 \Rightarrow \frac{1}{ab} < 1 $.
In addition since $ \kappa \left( A \right) = \frac{{\alpha}_{1}}{{\alpha}_{2}} $ and $ \det \left( A \right) = {\alpha}_{1} {\alpha}_{2} $ it follows that $ \det \left( A \right) \kappa \left( A \right) = {\alpha}_{1}^{2} $ and $ \det \left( A \right) {\det \left( B \right)}^{2} \kappa \left( C \right) = {\gamma}_{1}^{2} $.
Assuming the inequality holds:
$$ \kappa \left( C \right) \leq \kappa \left( A \right) \Rightarrow \det \left( A \right) \kappa \left( C \right) \leq \det \left( A \right) \kappa \left( A \right) $$
Since $ \det \left( A \right) > 0 $, now this implies:
$$ \frac{{\gamma}_{1}^{2}}{\det \left( B \right) \det \left( B \right)} \leq {\alpha}_{1}^{2} \Rightarrow {\gamma}_{1} \leq \det \left( B \right) {\alpha}_{1} $$
Namely, showing the above inequality of the Eigen Values is equivalent of proving the original inequality (Because all terms are positive).
One could see, using the Characteristic Polynomial of the matrix $ A $ and $ C $ that $ {\alpha}_{1} = \frac{a + b}{2} + \frac{\sqrt{{\left( a - b \right)}^{2} + 4}}{2}, \; {\gamma}_{1} = 1 + \frac{1}{\sqrt{ab}} $.
Working on the right hand of the inequality:
$$
\begin{align}
\det \left( B \right) {\alpha}_{1} & = \frac{1}{\sqrt{a b}} \left( \frac{a + b}{2} + \frac{\sqrt{{\left( a - b \right)}^{2} + 4}}{2} \right) \\
& = \frac{a + b}{2 \sqrt{a b}} + \sqrt{\frac{{\left( a - b \right)}^{2} + 4}{4 a b}} \\
& \geq 1 + \sqrt{\frac{{\left( a - b \right)}^{2} + 4}{4 a b}} \\ 
& \geq 1 + \sqrt{\frac{4}{4 a b}} \\
& = {\gamma}_{1}
\end{align}
$$
Where the first inequality comes from the Inequality of Arithmetic and Geometric Means and the second by removing a non negative term.
