This is example 16 in Chapter 2, Section 5 of Spanier's Algebraic Topology. It states that: Any infinite product of $1$-sphere has no universal covering space.
A universal covering space, as far as I know, need not to be simply connected. One counter example is offered in the same section of the book as Example 18. Any help would be appreciated.
The definition in Spanier for a universal cover is:
If $p \colon \tilde{X} \to X$ is a universal cover. Then for any covering projection $q \colon \tilde{X}^\prime \to X$, there is $f \colon \tilde{X} \to \tilde{X}^\prime$ such that $q \circ f = p$.
I do agree that a simply connected covering space of $X$ is an universal covering space of $X$, but I don't think the converse hold.
Best Answer
Another argument using the free action of $\pi_1$ on the fibers of a universal cover can be given as follows:
Suppose $Y:=\prod_{i=1}^\infty S^1$ has a universal covering space $p:X\rightarrow Y$. Fix $x_0\in S^1$. Consider the element $\overline x=(x_0,\ldots,x_0,\ldots)$ of $Y$. There is an open nhood $U$ of $\overline x$ and some open $\tilde U$ of $X$, such that $p\upharpoonright\tilde U$ is a homeomorphism onto $U$. We may assume $U$ has the form $$\prod_{i=1}^{n-1}U_i\times\prod_{i=n}^\infty S^1,$$ where each $U_i$ is an open nhood of $x_0$. Now consider the loop $\alpha:S^1\rightarrow \prod_{i=0}^\infty S^1$ given by $x\mapsto (x_0,\ldots,x_0,x,x_0,\ldots)$, where the $``x"$ appears in the $n$th position. Then $\alpha$ is a loop of $U$. As $p\upharpoonright\tilde U$ is a homeomorphism we get that $\alpha$ lifts via $p$ to a loop $\gamma$ of $X$ contained in $\tilde U$.
However, $\alpha$ is a non-trivial loop of $Y$, and thus as $p:X\rightarrow Y$ is the universal cover of $Y$, we must have that the lift of $\alpha$ is not a loop.
This contradiction shows $Y$ cannot have a universal covering.