⚡ AZUCATION SHORTCUT:
For $[x^2] = [x]^2$ where $[x]=n$, we must have $n^2 \le x^2 < n^2+1$.
This gives the general solution: $x \in [n, \sqrt{n^2+1})$.
Just test $n=3$ and $n=5$ to quickly identify Choice 4.

Step 1: Property of Greatest Integer Function
Let $[x] = n$, where $n$ is an integer. Then $n \le x < n+1$.
The equation is $[x^2] = n^2$.
By the definition of GIF, this implies: $n^2 \le x^2 < n^2 + 1$.
Taking the square root: $n \le x < \sqrt{n^2 + 1}$.

Step 2: Testing values in range $3 \le x \le 6$
* Case $n=3$: $3 \le x < \sqrt{3^2+1} \implies x \in [3, \sqrt{10})$.
* Case $n=4$: $4 \le x < \sqrt{4^2+1} \implies x \in [4, \sqrt{17})$.
* Case $n=5$: $5 \le x < \sqrt{5^2+1} \implies x \in [5, \sqrt{26})$.
* Case $x=6$: $[6^2] = 36$ and $[6]^2 = 36$. (True)

Step 3: Finding the Subset
The complete set $S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}$.
Choice 4 is $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$.
Since $(3, \sqrt{10}) \subset [3, \sqrt{10})$ and all other parts are within $S$, Choice 4 is a valid subset.

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