Working with the usual topology in $\mathbb{R}$
My main approach is to use induction principle. For $\Sigma_{1}^{0}$ I used the density of rationals in the reals, so inside an open interval, lets say (a,b) I can decrease the intervals infinitely into one smaller.
My induction hypothesis is cardinality of $\Sigma_{\alpha}^{0}$ is $2^{\aleph_{0}}$, but I dont know how can I prove $\Sigma_{\alpha +1 }^{0}$ cardinality is $2^{\aleph_{0}}$
Given that $\Sigma^0_\alpha$ has the cardinality $2^{\aleph_0}$ of the continuum, you immediately get the same for $\Pi^0_\alpha$. The number of countable sequences of $\Pi^0_\alpha$ sets is therefore $\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0}$. Every $\Sigma^0_{\alpha+1}$ set is the union of such a sequence, so there are at most $2^{\aleph_0}$ $\Sigma^0_{\alpha+1}$ sets. (You already know that there are at least that many.)
For limit ordinals $\alpha<\omega_1$, the class $\Sigma^0_\alpha$ is the union of countably many classes $\Sigma^0_\beta$ for $\beta<\alpha$, each of cardinality at most $2^{\aleph_0}$, so you get at most $2^{\aleph_0}$ $\Sigma^0_\alpha$ sets.
Although it wasn't in your question, the class of all Borel sets is the union of the $\aleph_1$ classes $\Sigma^0_\alpha$ for all $\alpha<\omega_1$, so it also has cardinality $\aleph_1\cdot2^{\aleph_0}=2^{\aleph_0}$.