1. We are given $f(x)=x^2,\, x\in\mathbb R$ and for any subset $A\subseteq \mathbb R$, $g(A)=\{x\in\mathbb R: f(x)\in A\}=\{x\in\mathbb R: x^2\in A\}.$

Also, $S=[0,4].$ We must determine which statement is not true.

2. First compute $f(S)$.

Since $S=[0,4]$ and $f(x)=x^2$, we get $f(S)=\{x^2:x\in[0,4]\}=[0,16].$

3. Now compute $g(S)$.

By definition, $g(S)=\{x\in\mathbb R:x^2\in[0,4]\}.$ This means $0\le x^2\le 4 \implies |x|\le 2.$ Hence, $g(S)=[-2,2].$

4. Compute $g(f(S))$.

Since $f(S)=[0,16]$, $g(f(S))=\{x\in\mathbb R:x^2\in[0,16]\}.$ So, $0\le x^2\le 16 \implies |x|\le 4.$ Therefore, $g(f(S))=[-4,4].$

5. Compute $f(g(S))$.

We found $g(S)=[-2,2]$. Then $f(g(S))=f([-2,2])=\{x^2:x\in[-2,2]\}=[0,4].$ Thus, $f(g(S))=S.$ So option B is true.

6. Check option C: $f(g(S))\subseteq f(S)$.

We have $f(g(S))=[0,4],\qquad f(S)=[0,16].$ Clearly, $[0,4]\subseteq[0,16].$ So option C is true.

7. Check option D: $g(f(S))=g(S)$.

We have $g(f(S))=[-4,4],\qquad g(S)=[-2,2].$ These are not equal. So option D is false.

8. Check option A. The printed symbol appears as "$e$", which from context is intended to mean either "$\subseteq$" or "$\in$". Let us check both possibilities:

• If option A means $g(f(S))\subseteq S,$ then $[-4,4]\subseteq[0,4]$ is false.

• If option A means $g(f(S))\in S,$ then this is also false, because $g(f(S))=[-4,4]$ is a set, not a real number in $[0,4]$.

So as written, A is also not true.

9. Conclusion.

From the actual set computations:

• $B$ is true,
• $C$ is true,
• $D$ is false,
• and $A$ as printed is also false.

Hence the question statement has an issue in option A. Among the standard intended options, the clearly false set-identity statement is $\boxed{D}.$