I am trying to prove:
Lee, Smooth Manifolds, Exercise 2.9. Show that an open cover $\{U_\alpha\}$ of $X$ is locally finite if and only if each $U\alpha$ intersects $U_\beta$ for only finitely many $\beta$.
However, on my way I found what I believe to be the counterexample: Let $X=\mathbb{R}$ and $\{U_\alpha\}=\{(k,k+2):k \in \mathbb{Z}\} \cup \{\mathbb{R}\}$. Then clearly $\{U_\alpha\}$ is locally finite since each point $p=k+\delta \in \mathbb{R}, k \in \mathbb{Z}, \delta \in [0,1)$ has a neighbourhood $(p-1,p+1)$ that intersects finitely many (3) sets from $\{U_\alpha\}$, notably $(k-1,k+1),(k,k+2),\mathbb{R}$. However, take $U_\alpha=\mathbb{R}$, then it intersects with $U_\beta=(k,k+2)$ for each $k \in \mathbb{Z}$, therefore for infinitely many $\beta$ which is a contradiction.
Is the cover by definition minimal? I.e. no proper subset of the cover of $X$ is the cover for $X$? That we have to remove $\mathbb{R}$ from $\{U_\alpha\}$?
Something has gone wrong with your transcription of the problem. Looking in Lee's book, I can't find the question that you quote.
However, question 1.4 is close. Part (a) is to prove that if a cover has the intersection property described, then it is locally finite. Part (b) is to prove that the converse does not hold in general, which you have done. And part (c) is to show that the converse does hold with the additional assumption that the elements of the cover are all precompact.