Proof for optimal interval scheduling using a Greedy Approach
Let's consider two jobs in the sequence you obtained:


$A$, of weight $w_A$ begins at $t_0$ and finish at $t_0 + s_A$
$B$ coming just after $A$, of weight $w_B$ begins at $t_0 + s_A$ and finish at $t_0 + s_A + s_B$


If we compute only $K_{A, B}$ the contribution of $A$ and $B$ in $K = \sum_j w_j C_j$:

$K_{A, B} = w_A (t_0 + s_A) + w_B (t_0 + s_A + s_B)$

If A and B are inversed in the sequence, we have $K'_{A, B}$:

$K'_{A, B} = w_A (t_0 + s_A + s_B) + w_B (t_0 + s_B)$

The difference is:

$\Delta K_{A, B} = K'_{A, B} - K_{A, B}$

$= w_A s_B - w_B s_A$

$= (w_A/s_A - w_B/s_B) \times (s_A s_B)$

The switch should be done if and only if $\Delta K_{A, B}$ is negative in order to minimize $K$. Only the approach 3 provides you a sequence where no more switch is worth.

If two jobs have the same $w/s$ ratio, just take them in any order, the final $K$ would remain unchanged.