On the definition of local compactness INTRODUCTION
Definition 1
A topological space $(S, {\cal{T}})$ is compact if any of its open covers contains a finite subcover.
Definition 2
In a topological space $(S, {\cal{T}})$, a set $ A\subset S$ is compact if every open cover of $ A$ contains a finite subcover.
Equivalently, $ A\subset S $ is compact if it is, with the subspace topology, a compact topological space.
Definition 3
A topological space $(S, {\cal{T}})$ is locally compact if any point of $S$ has a compact neighbourhood, i.e. if for  $\forall x\in S$ there exist an open $U\in{\cal{T}}$ and a compact $A\subseteq S$ such that $x\in U\subseteq A$.
QUESTION 1
Consider an arbitrary open set $U$ from the topology $\cal T$.
Would it be right to say that it is part of any cover that covers it?
QUESTION 2
If the answer to Q 1 happens to be affirmative, then the said $U$ must be compact, because it serves as its own finite subcover.
Is this correct?
QUESTION 3
If the answer to Q 2 is affirmative also, then we are arriving at a paradoxical conclusion: since each point resides in some $U\in\cal T$, and since each such $U$ is compact, then every point has a compact neighbourhood --- and the topological space is always locally compact.
Where did I mess up? (In Q 1, I guess?)

The answer to your question $1$ is no. Consider the set $U:=(0,1)\subset \mathbb R$ equipped with the usual topology. The collection of open sets $\{(1/n,1-1/n)\}_{n\in\mathbb N}$ covers $U$, but $U$ is not in the collection. $U$ is definitely not compact.
