I want to prove, by applying Brun's sieve, that $$\pi(x)\ll\dfrac{x}{\log x}$$ By setting $\mathscr{A}=\left\{\,n\le x\right\} $ and by Brun's sieve, to sift $\mathscr A$ for all numbers less than or equal to $x$ that is divisible by prime $p\le z$, which $z$ is going to be $\sqrt[100]{\dfrac{x}{\log x}}\ll\dfrac{x}{\log x}$ and $\log z\asymp\log x$. Consequently, $\omega(p)=1$ and $\kappa =1$ satisfying the condition
$i) \text{ } 0\le\dfrac{\omega(p)}{p}\le \dfrac{1}{2}$
$ii) \text{ }$ $\displaystyle \sum_{w\le p\le z}\dfrac{\omega(p)\log p}{p}=\sum_{w\le p\le z}\dfrac{\log p}{p}\le \log\dfrac{z}{w}+\text{const}=\kappa\log\dfrac{z}{w}+\text{const}$
Then choosing $\lambda$ to be a number that satisfies $0<\lambda e^{1+\lambda}<1$ together with selecting $b=1$. we have that $\pi(x)\le z+S(\mathscr A,\mathscr P, z)$, where $S(\mathscr A,\mathscr P, z)$ is the number of the elements in $\mathscr A$ left from being sieved.
Then by Bruns' theorem $$S(\mathscr A,\mathscr P, z)\le x\prod_{p\le z}\left(1-\dfrac{1}{p}\right)\left(1+2\cdot\dfrac{\lambda^3e^{2\lambda}}{1-\lambda^2e^{2+2\lambda}}\exp{\left(\dfrac{c_1}{\lambda\log z}\right)}\right)+O\left(z^{2+\frac{2.01}{e^{2\lambda}-1}}\right)$$ And by Merten's Theorem $\text{ }\displaystyle\prod_{p\le z}\left(1-\dfrac{1}{p}\right)=\dfrac{e^{-\gamma}+o(1)}{\log z}$ with $2\cdot\dfrac{\lambda^3e^{2\lambda}}{1-\lambda^2e^{2+2\lambda}}\exp{\left(\dfrac{c_1}{\lambda\log z}\right)}\longrightarrow \text{const}$. The error term above and $z$ $\ll \dfrac{x}{\log x}$ as well. Therefore, we have $\pi(x)\ll \dfrac{x}{\log x}$
I'm not sure that I choose $z$ and use the sieve appropriately to tackle the problem.