Let $X$ be locally path-connected, path-connected, and semilocally simply connected by paths. Suppose that $X$ admits a covering with $n$ sheets for every $n \in \mathbb{N}$. Show then that $X$ admits a covering with infinitely many sheets.
I do this:
We know that $X$ admits a universal covering that covers all the covering with $n$ sheets for every $n \in \mathbb{N}$, and so the subgroup corresponding to the universal covering (i.e., the trivial one) has an index greater than the indices of the subgroups corresponding to coverings with $n$ sheets, which have an index of $n$. However, this implies that the subgroup corresponding to the universal covering has an infinite index, and therefore the universal covering has an infinite number of sheets.
i want to know if the demonstration is good
The argument looks fine.
Technically I guess that when the $n$-fold covering factors through the universal covering, that shows that the degree of the universal covering is $\ge n$, not $> n$ as you suggest. But the argument works the same.