It is said for the above question the answer is Option A) A horizontal line between $x_1$ and $x_1-1$.
I am unable to understand how the likelihood is being expressed in terms of $x_1$ rather than it being constant and equal to one for $\theta < x_1 ≤ \theta+1$ and $0$ otherwise.
We're looking for $\theta$ given the sample data $x_1$ and given the parameterization of the distribution, all the possible/likely candidates for $\theta$ are at the interval $(x_1-1, x_1)$
So the sample $x_1$ can only come from this interval $$\theta<x_1<\theta+1$$ Which is just two inequalities
$$\theta<x_1$$
$$x_1<\theta+1\Rightarrow x_1-1<\theta$$
Now just rearranging the inequalities to isolate $\theta$ we would have $$x_1-1<\theta<x_1$$
Intuition example:
Say the actual value of $\theta$ is $3.5$. So your true uniform distribution is on the interval $(\theta,\theta+1)=(3.5,4.5)$ (in red).
Suppose you did not know this true $\theta$ and the only information you have is a sample from the true distribution and suppose this sample is $4$ (in blue).
Your best bet to guess the starting point of the true distribution is to look at the starting direction which is to the left of the sample point. You no longer care about the end point because the "length of one" restriction makes it so that knowing the start already gives "sufficient" information about the end point.
So $(x_1-1,x_1)$ captures the true value of $\theta$. Thinking of extreme cases, $x_1$ might be the true end point but still the interval $(x_1-1,x_1)$ would capture the true value of $\theta$.
In a more advance but related topic, this restricted distribution is considered to belong to a curve exponential family where in the parameter space is restricted wherein knowing one parameter already fixes the value of the other parameters of the distribution, hence the only need to guess a single parameter.