Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$, where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$, then a possible subset of $S$ is

$[x]$ denotes the greatest integer not exceeding $x$. For example, $[3.5] = 3$, $[4] = 4$, $[5.99] = 5$.

Since $3 \le x \le 6$, the possible values of $[x]$ are $3, 4, 5, 6$. Each case is considered separately.

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When $[x] = 3$, i.e., $3 \le x < 4$:

$[x]^2 = 9$

For $[x^2] = 9$, we need $9 \le x^2 < 10$

$3 \le x < \sqrt{10}$

Valid range: $[3, \sqrt{10})$, i.e., $[3, 3.162...)$

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When $[x] = 4$, i.e., $4 \le x < 5$:

$[x]^2 = 16$

For $[x^2] = 16$, we need $16 \le x^2 < 17$

$4 \le x < \sqrt{17}$

Valid range: $[4, \sqrt{17})$, i.e., $[4, 4.123...)$

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When $[x] = 5$, i.e., $5 \le x < 6$:

$[x]^2 = 25$

For $[x^2] = 25$, we need $25 \le x^2 < 26$

$5 \le x < \sqrt{26}$

Valid range: $[5, \sqrt{26})$, i.e., $[5, 5.099...)$

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When $[x] = 6$, i.e., $x = 6$:

$[x]^2 = 36$ and $[x^2] = [36] = 36$

Valid value: $\{6\}$

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Combining all cases:

$S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}$

Any valid subset of $S$ must have all its elements lying within these intervals. Values like $3.5, 4.5, 5.5$ do not belong to $S$ since they fall outside the valid ranges.

The answer is Option 3.