Example 13 Given that the function $f(x)$ is defined on $[0,1]$. And it satisfies the following conditions: (1) $f(1)=3$; (2) $f(x) \geqslant 2$; (3) If $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{1}+x_{2} \leqslant 1$, then $f\left(x_{1}+x_{2}\right) \geqslant f\left(x_{1}\right)+f\left(x_{2}\right)-2$. (1) Find the maximum and minimum values of the function $f(x)$; (2) Compare $f\left(\frac{1}{2^{*}}\right)$ with $\frac{1}{2^{*}}+2$ $\left(n \in \mathbf{N}^{+}\right)$; (3) Someone noticed that when $x=\frac{1}{2^{n}}(n \in \mathbf{N})$, $f(x)<2 x+2$. Based on this, he proposed a conjecture: For all $x \in(0,1]$, $f(x)<2 x+2$. Please determine whether the conjecture is correct and explain your reasoning.
(1) Omitted.
(2) From (1), it is easy to know that $f(x)$ is an increasing function. Let $x_{1}=x_{2}=\frac{1}{2^{n+1}}$, then $x_{1}+x_{2}=\frac{1}{2^{n}}$, so $f\left(\frac{1}{2^{n}}\right) \geqslant f\left(\frac{1}{2^{n+1}}\right)+f\left(\frac{1}{2^{n+1}}\right)-2$, thus $f\left(\frac{1}{2^{n}}\right)-2 \geqslant 2\left[f\left(\frac{1}{2^{n+1}}\right)-2\right]$. Let $a_{n}=f\left(\frac{1}{2^{n}}\right)-2$. Then $a_{n} \geqslant 2 a_{n+1}$, so $a_{n} \leqslant \frac{1}{2} a_{n}(n \geqslant 1)$, which means $a_{n} \leqslant\left(\frac{1}{2}\right)^{n}[f(1)-2]=\left(\frac{1}{2}\right)^{n}$; thus $f\left(\frac{1}{2^{n}}\right)-2 \leqslant\left(\frac{1}{2}\right)^{n}$, so $f\left(\frac{1}{2^{n}}\right) \leqslant 2+\left(\frac{1}{2}\right)^{n}$.
(3) For any $x \in(0,1]$, there always exists $n \in \mathbf{N}^{+}$, such that $\frac{1}{2^{n-1}}2 \cdot \frac{1}{2^{n 11}}+2=\frac{1}{2^{\prime \prime}}+2$. Therefore, $f(x)<2 x+2$.