Let $S \subset \mathbb{N}$ with each number $n \in S$ with probability $1/2$ independently; in other words we run through all $n \in \mathbb{N}$ and each time $n$ becomes part of $S$ with chance $1/2$ independently.Prove that
$$\mathbb{P}\bigg[ \lim_{n \rightarrow \infty} \frac{\lvert S \cap [n] \rvert}{n} = \frac{1}{2} \bigg] = 1$$
Hint: For fixed $\varepsilon > 0$ and $n$ use the Chernoff's bound to find an upper bound for
$$\mathbb{P}\bigg[ \bigg\lvert \frac{\lvert S \cap [n] \rvert}{n} -\frac{1}{2} \bigg\rvert \ge \varepsilon\bigg]$$
and apply union bound to show that
$$\mathbb{P}\bigg[ \exists n \ge n_0 \text{ with } \bigg\lvert \frac{\lvert S \cap [n] \rvert}{n} -\frac{1}{2} \bigg\rvert \ge \varepsilon\bigg] \overset{N \rightarrow \infty}{=} \mathcal{o}(1) $$
Does this also work if we use Chebyshev's bound?
Remark: $[n]$ denotes the set $\{1,2,\ldots,n\}$.
First we note that $\lvert S \cap [n] \rvert = \sum_{k=1}^n \mathbf{1}_{\{1\}}(X_k)$ for $X_k \sim Ber(1/2)$ i.i.d. Thus for $Y_n := \frac{\lvert S \cap [n] \rvert}{n}$ we have $\mathbb{E}[Y_n] = 1/2$ and $\mathbb{V}[Y_n] = 1/4$.
$$\mathbb{P}\bigg[\bigg\lvert X - \mathbb{E}[X] \bigg\rvert \ge \varepsilon\bigg]=\mathbb{P}\bigg[\bigg\lvert \frac{\lvert S \cap [n] \rvert}{n} - \frac{1}{2} \bigg\rvert \ge \varepsilon\bigg] \le \exp\bigg(-\frac{\varepsilon^2}{2(1/2 + \varepsilon/3)} \bigg)$$
However, when I try to apply the union bound I get
\begin{align} \mathbb{P}\bigg[ \exists n \ge N \text{ with } \bigg\lvert \frac{\lvert S \cap [n] \rvert}{n} -\frac{1}{2} \bigg\rvert \ge \varepsilon\bigg] &\le \sum_{k \ge N}^\infty \mathbb{P}\bigg[\bigg\lvert Y_k -\frac{1}{2}\bigg\rvert \ge \varepsilon \bigg] \\ &= \sum_{k \ge N}^\infty \exp\bigg(-\frac{\varepsilon^2}{2(1/2 + \varepsilon/3)} \bigg) \end{align}