Is every probability measure in the line induced by a random variable?

Yes, trivially: consider the probability space $\Omega = \mathbb{R}$, $\mathcal{F} = \mathcal{B}$, $\mathbb{P} = \mu$, and let $X : \Omega = \mathbb{R} \to \mathbb{R}$ be the identity map $X(\omega) = \omega$.

Alternatively, use inverse transform sampling. Let $F(x) = \mu((-\infty, x])$ be the cumulative distribution function of $\mu$, and let $G(t) = \inf\{x : F(x) \ge t\}$ be its "inverse". (If $F$ is actually 1-1 then $G$ is truly the inverse of $F$.) Now let $U$ be a Uniform(0,1) random variable on any probability space $(\Omega, \mathcal{F}, \mathbb{P})$ (e.g. the identity map on $[0,1]$ with Lebesgue measure) and set $X = G(U)$. It is then easy to see that the measure induced by $X$ is again $\mu$. Proof: we have $\mathbb{P}(X \le x) = \mathbb{P}(G(U) \le x) = \mathbb{P}(U \le F(x)) = F(x)$. So the measure induced by $X$ has $F$ as its cdf, and this uniquely determines it as being $\mu$.

Yes. Let $(\Omega ,\mathcal F,\mathbb P) = ([0,1],\mathcal B([0,1]),\mathcal L)$ where $\mathcal L$ is the Lebesgue measure on $[0,1]$. Fix your measure $\mu$, and write $F(x) = \mu((-\infty,x])$.

Let $X(\omega)= \inf\{z : F(z) \ge \omega \}$. This is known as the Skorokhod Representation of the random variable $X$.

I'll leave you to verify that the distribution of $X$ is exactly given by $F$. (However, if this gives you trouble, see section 3.12 of the reference below.)

Reference:

Williams, D. (1991). Probability with martingales. Cambridge university press.