homotopy equivalence of these three spaces

You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).

I shall use the following two well-known theorems:

If $A \hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X \to X/A$ is a homotopy equivalence.

If $X$ is a CW-complex and $A$ is a subcomplex, then $A \hookrightarrow X$ is a cofibration.

Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $\mathbb{S}^{2}/\mathbb{S}^{0}$ is homotopy equivalent to $\mathbb{S}^{2} \vee \mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.

To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:

Let $f : X \to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.

Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.