Chọn phương án B.

Điều kiện: \(\begin{cases}
x+3>0\\ x>0
\end{cases}\Leftrightarrow x>0\).

Ta có $$\begin{aligned}
&\,\log_{\sqrt{2}}(x+3)-\log_2x\leq4\\
\Leftrightarrow&\,\log_{\sqrt{2}}(x+3)\leq\log_2x+4\\
\Leftrightarrow&\,\log_{2^{\frac{1}{2}}}(x+3)\leq\log_2x+\log_22^4\\
\Leftrightarrow&\,2\log_{2}(x+3)\leq\log_216x\\
\Leftrightarrow&\,\log_{2}(x+3)^2\leq\log_216x\\
\Leftrightarrow&\,(x+3)^2\leq16x\\
\Leftrightarrow&\,x^2-10x+9\leq0.
\end{aligned}$$

Suy ra \(x\in[1;9]\).

Vậy \(S=[1;9]\).