Let $X=\{ x_{1},x_{2},x_{3},x_{4} \}$.
Let $\mathcal{B}= \{ \{ x_{1}\}, \{ x_{3}\} , \{ x_{1},x_{2},x_{3} \} , \{ x_{1},x_{3},x_{4} \} \}$ be a basis for a topology on $X$.
I am trying to find a covering space of this space other than $X$ itself.
I was thinking about taking a bouquet of 4 circles, and mapping each circle to one of the points. Would this work as a covering map? I'm not sure it would be a homeomorphism when restricted.
Or instead maybe define a map from each circle into one of the elements of the basis, since this would allow each point in $X$ to have an evenly covered neighbourhood, but I cannot see how to define this map explicitly.
If you have any covering map $p : Y \to X$, then the fibers $p^{-1}(x)$ are discrete. Thus your suggestion does not work. However, if $F$ is any discrete space, then the projection $\pi : X \times F \to X$ is always a covering map.
For $F = \mathbb Z$ this resembles მამუკა ჯიბლაძე's example. A basis of open sets of $X \times F$ is given by the $B \times \{n\}$, where $B \in \mathcal B$ and $n \in \mathbb Z$. Now identify the set $X \times \mathbb Z$ with $\mathbb Z$ via the bijection $f(x_i,n) = 4n+i$. Then $f(\{x_1 \} \times \{n\}) = \{4n+1 \}$, $f(\{x_3 \} \times \{n\}) = \{4n+3 \}$, $f(\{x_1 \} \times \{n\}) = \{4n+1 \}$, $f(\{x_1,x_2,x_3 \} \times \{n\}) = \{4n+1,4n+2,4n+3 \}$, $f(\{x_1,x_3,x_4 \} \times \{n\}) = \{4n+1,4n+3,4n+4 \}$.
Note that Mamuka Jibladze took $\{4n+3,4n+4,4n+5 \}$ instead of the last set.