This is not intended as a full answer but as a series of hints. First, look at your likelihood.

We have

$$L(\theta) = \prod_{i=1}^n 1 \quad I\left( \theta \leq x_i \leq \theta + 1 \right)$$

where $I( x \in A)$ is the indicator function, returning $1$ if indeed $x \in A$ and zero otherwise. This is how the parameter $\theta$ enters our likelihood here.

Note here that for the likelihood to be non-zero, all the $x$s have to be in that interval. But then, this will happen if and only if the minimum is greater than $\theta$ and the maximum is smaller than $\theta+1$, would you agree? These conditions will give you an interval in which $\theta$ can lie.

But since this is an interval and the likelihood is constant in that interval, the mle is not unique in this case. So if you are looking at this problem by yourself, you have the freedom of choice. However, only one of these three estimators fullfils the condition. I leave it to you to find which one.

This is not intended as a full answer but as a series of hints. First, look at your likelihood.

We have

$$L(\theta) = \prod_{i=1}^n 1 \quad I\left( \theta \leq x_i \leq \theta + 1 \right)$$

where $I( x \in A)$ is the indicator function, returning $1$ if indeed $x \in A$ and zero otherwise. This is how the parameter $\theta$ enters our likelihood here.

Note here that for the likelihood to be non-zero, all the $x$s have to be in that interval. But then, this will happen if and only if the minimum is greater than $\theta$ and the maximum is smaller than $\theta+1$, would you agree? These conditions will give you an interval in which $\theta$ can lie.

But since this is an interval and the likelihood is constant in that interval, the mle is not unique. So if you are looking at this problem by yourself, you have the freedom of choice. That is, you can take a boundary point or a point in the interior, it does not matter. However, only one of these three estimators fullfils the condition. I leave it to you to find which one.