Suppose we partition the powers of $2$ into two sets in the following manner: $$A = \{1\}, \quad B = \{2,4,8,16,\dots\}$$ Now suppose that, for each odd positive integer $t$, we select either $tA$ or $tB$, and form the union $S$ of all sets selected in this way. For example, if we always choose $tA$, then the resulting set is $$S_1 = 1A\cup 3A\cup 5A\cup\cdots = \{1,3,5,7,9,\dots\},$$ while if we always choose $tB$, the resulting set is $$S_2 = 1B\cup 3B\cup 5B\cup\cdots = \{2,4,6,8,10,\dots\}.$$ These sets consist of all odd and all even integers, respectively; they each have natural density $\frac 12$.
On the other hand, we might form a "mixed" union. A tame example is \begin{align*} S_3 = \left(\bigcup\limits_{t\equiv 1\text{ mod }4} tA\right) \cup \left(\bigcup\limits_{t\equiv 3\text{ mod }4} tB\right) &= 1A\cup 3B\cup 5A\cup 7B\cup\cdots \\ &= \{1,5,6,9,12,13,14,\dots\}, \end{align*} which again can be shown to have natural density $\frac 12$. A more wild example is \begin{align*} S_4 = \left(\bigcup\limits_{\substack{t\text{ odd} \\ \lfloor\log_2 t\rfloor\text{ even}}} tA\right) \cup \left(\bigcup\limits_{\substack{t\text{ odd} \\ \lfloor\log_2 t\rfloor\text{ odd}}} tB\right) &= 1A\cup 3B\cup (5A\cup 7A)\cup (9B\cup\cdots\cup 15B)\cup\cdots \\ &= \{1,5,6,7,12,17,18,19,21,22,23,24,\dots\}. \end{align*} This set has no natural density; in particular, $\frac{|S_4\cap\{1,2,\ldots,n\}|}{n}$ oscillates between $\frac 7{18}$ (when $n=2^{2k}$) and $\frac{11}{18}$ (when $n=2^{2k+1}$).
My question: Is it true that, no matter how we build $S$, its lower (resp., upper) asymptotic density will be less than or equal to (resp., greater than or equal to) $\frac 12$? This strikes me as a very reasonable conjecture, but I cannot figure out how to prove it.
We can prove the conjecture true by first considering the logarithmic density of a set $T$ of positive integers, which is $$ \lim_{x\to\infty} \frac1{\log x} \sum_{m\in T\cap[1,x]} \frac1m. $$
Proposition: Any set $S$ formed as described in the OP has logarithmic density $\frac12$.
Proof: Let $x\ge1$, and consider any odd integer $t\le x$. Note that $\displaystyle \sum_{m\in tA\cap[1,x]} \frac1m = \frac1t, $ while $$ \sum_{m\in tB\cap[1,x]} \frac1m = \frac1t -\frac1{2^kt} $$ where $k$ is the largest nonnegative integer with $2^kt\le x$. Consequently, whether $C_t=A$ or $C_t=B$, we have $$ \sum_{m\in tC_t\cap[1,x]} \frac1m = \frac1t +O\biggl( \frac1x \biggr). $$ It follows that $$ \sum_{m\in S\cap[1,x]} \frac1m = \sum_{\substack{t\le x \\ t\text{ odd}}} \sum_{m\in tC_t\cap[1,x]} \frac1m = \sum_{\substack{t\le x \\ t\text{ odd}}} \biggl( \frac1t +O\biggl( \frac1x \biggr) \biggr) = \frac12\log x + O(1), $$ which implies the proposition. $\quad\square$
While the existence of a set's logarithmic density does not imply the existence of its natural density, it does imply bounds on the set's upper and lower natural densities. Suppose, for example, that the lower natural density of $S$ exceeds $\frac12$. This implies that there is a constant $\alpha>\frac12$ such that $\#(S\cap[1,x]) > \alpha x$ when $x$ is sufficiently large, say $x>x_0$. It follows by partial summation that we would have \begin{align*} \sum_{m\in S\cap[1,x]} \frac1m &= \int_{1-}^x \frac1u \,d\#(S\cap[1,u]) \\ &= \frac{\#(S\cap[1,x])}x - 0 + \int_1^x \frac{\#(S\cap[1,u])}{u^2}\,du \\ &> O(1) + \int_{x_0}^x \frac\alpha u\,du = \alpha\log x+O(1), \end{align*} which would contradict the logarithmic density of $S$ equaling $\frac12$. An analogous argument works for upper densities.
(In the last calculation, the second expression is a Riemann–Stieltjes integral, and the third expression follows from it via integration by parts; however, one can directly verify that the first and third expressions are equal by dividing the domain of integration in the third expression into subintervals on which $\#(S\cap[1,u])$ is constant, and then integrating and simplifying the resulting telescopic sum.)