Suppose that $\sqrt 3$ is rational. $\Rightarrow \sqrt 3 = \frac {p}{q}$ where $p$, $q$ are in $\mathbb Z$, $q\ne0$
$\Rightarrow p=q\sqrt 3$.
We now consider the set $ S=\{k\sqrt 3: k, k\sqrt 3\in \mathbb Z^+\}$
By our supposition, $S$ is a nonempty set of positive integers which by the WOP, has a smallest member $s$, say, and has the form $s=t\sqrt 3$ for some integer $t$.
Now $s\sqrt 3-s=s\sqrt 3-t\sqrt 3=(s-t)\sqrt 3$
But $s\sqrt 3=t\sqrt 3\sqrt 3=3t$ where $s$, $t$ are integers.
$\Rightarrow 3t-s=(s-t)\sqrt 3$ where $s$, $t$ are integers.
$\Rightarrow (s-t)\sqrt 3$ is an integer
which is positive as $s-t=t\sqrt 3-t=t(\sqrt 3-1)$ and $\sqrt 3\gt 1$
i.e., $(s-t)\sqrt 3\in\mathbb Z^+$.
However, $s(\sqrt 3-1)\lt s$ as $\sqrt 3-1\lt 1$.
But this contradicts the definition of $s$ as the smallest element in $S$.
Hence, the supposition that $\sqrt 3$ is rational is false.
The above is the proof given by the textbook. I doubt why the smallest element of $S$ must have the form of $t\sqrt 3$. Let's say $\sqrt 3 = \frac {p}{q}$, where $p$, $q$ are in the simplest form, so $p$, $q$ have no common factors. Then $q$ will be the smallest element of $S$, but cannot be expressed as $t\sqrt 3$. How do you justify this?
First let's start with the other classical proof of $\sqrt 3$ irrationality. (Note that it is short because it uses powerful theorem on prime numbers).
If $\sqrt 3=\frac pq$ with $\gcd(p,q)=1$ then $p^2=3q^3$ so $3\mid p^2$ but $3$ being a prime it also divide $p$.
So $p=3k$ and $p^2=9k^2=3q^2\Rightarrow3k^2=q^2\Rightarrow3\mid q^2$ but $3$ being a prime it also divide $q$.
This is a contradiction of $\gcd(p,q)=1$, so $\sqrt 3$ should be irrationnal.
Your proof is globally ok, but at first reading I found it a little confusing, especially about this kind of duality on wether you consider $t$ or $s$ for the minimality.
I think it would be clearer if you define $S=\{k\in\mathbb N^*\mid k\sqrt 3\in\mathbb Z^+\}$ and claim that $\exists\ k_0=\min(S)$.
Using the same letter $k$ used to define $S$ and indice $0$ for minimum greatly improve the significance of the naming and readability.
$s_0=k_0\sqrt 3$
Let's define $s=(s_0-k_0)\sqrt 3\quad$ and then by your proof we get that $k=s_0-k_0\in S\ $ ($k$ is positive and $s$ is an integer, this part is ok).
This is then the part of your proof I find a little confusing, I propose the following composition instead :
$k-k_0=s_0-k_0-k_0=k_0\sqrt 3-2k_0=\underbrace{(\sqrt 3-2)}_{<0}\underbrace{k_0}_{>0}<0$
Thus $k<k_0$ contradicting the minimality of $k_0$.
Regarding your other question of why $q$ cannot be expressed as any $t\sqrt 3$, I return you the question "why not ?". You are making the supposition that $\sqrt 3$ is rationnal after all.