Is there a probabilistic proof of the inequality $4p(1-p) \leq 1$ for a probability $p$?

A (possibly unsatisfying) solution:

Let X be a binomial random variable with 2 trials and probability of success $|1-2p|$. Then the probability that $X \neq 2$ is $1-|1-2p|^2 = 1 - (1 - 4p + 4p^2) = 4p - 4p^2 = 4p(1-p)$.

In terms of your example, let $X_1$ and $X_2$ be i.i.d. Bernoulli variables with probability $|1-2p|$. Then $4p(1-p)$ is the probability that the two $X_i$ are not both 1.

Alternative solution:

If $p \leq \frac{1}{2}$, then let be a binomial random variable with 2 trials and probability of success 2p. Then $P(X > 0) = 1 - (1-2p)^2 = 1 - (1 - 4p + 4p^2) = 4p - 4p^2 = 4p(1-p)$.

If $p \geq \frac{1}{2}$, then let $q = 1-p$ and let X be a binomial random variable with 2 trials and probability of success 2q. Then $P(X > 0) = 1 - (1-2q)^2 = 4q(1-q) = 4(1-p)p$.