How the Ornstein–Uhlenbeck process can be considered as the continuous-time analogue of the discrete-time AR(1) process?

In case $p=1$ you have $$ x_{k+1} = c+a x_k + b\varepsilon_k $$ so that if you put $c = \theta\mu\Delta t$, $a = -\theta\Delta t$ and $b = \sigma\sqrt{\Delta t}$ you get $$ x_{k+1} = \theta(\mu - x_k)\Delta t + \sigma \varepsilon_k\sqrt{\Delta t} $$ which is exactly an Euler-Maryuama discretization of OU at times $(k\Delta t)_{k\in \Bbb N_0}$.

Rather than using an approximation (like Euler-Maryuama), one can just sample* the continuous-time solution:

$$x(t)=x(0)\,e^{-\theta t}+\mu \,(1-e^{-\theta t})+\sigma \int _{0}^{t}e^{-\theta (t-s)}\,dW_{s}.\, $$

Writing $X_k=x(k\,\Delta t)$, the sampled process is described by the AR(1) process: $$ X_{k+1} = c + \varphi \,X_k + \varepsilon_k $$ where: \begin{align} X_0&=x(0),\\ \varphi &= e^{-\theta \Delta t},\\ c &= (1-\varphi)\mu, \\ \epsilon_k &\sim \mathcal{N}\Big(0,\frac{1}{2\theta}\sigma^2\big(1-e^{-2 \theta \Delta t}\big)\Big). \end{align}

_{* I mean sample, not sample}

Should be a comment on Ilya's answer but I don't have enough reputation --

Although the other answer does show how the AR process is equivalent to the OU process, keep in mind that the Euler-Maruyama discretization is just an approximation. In order to derive the exact relationship between the two models, you would have to integrate the OU process from time t to t+1, and then derive the various relationships between the parameters. Luckily there is a closed form solution to the OU SDE, so this is not too difficult.

EDIT: you could also use the Milstein Discretization, which is a more accurate approximation because it includes the convexity term from Ito's lemma. But it is still easier than the exact solution.