I need to prove that Hawaiian Earring has no universal cover. By 'universal' I mean all other covers are below it. For instance if $Y$ is a covering space of $X$ and $Z$ is an another covering space of $X$, then $Y$ is also covering space of $Z$. In this sense $Y$ is the universal covering space.
By Hawaiian Earring I mean $X=\bigcup_n C_n$ where $C_n$ is a circle with center $(1/n,0)$ and radius $1/n$, i.e. $C_n=\lbrace x,y \in \mathbb{R}^2 \: | \: (x-1/n)^2 +y^2 = 1/n^2 \rbrace $.
Let's define $g_k:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $g_k(x,y)= (\frac{x}{n^{(k+1)/n}}, \frac{y}{n^{(k+1)/n}})$. Ofc it's homeomorphism cuz it's just multiplication by a constant. Moreover I am able to prove that $g_k(C_t)=C_{t+k+1} $, thus $g_k(X)=g_k(\bigcup_n C_n)=\bigcup_{k+1} = X_k$ where $X_k=\bigcup_{n>k} C_n$. So $g_k (X)= X_k$.
Now I gonna to construct covering space of $X$ and then show that it ain't universal cover. But from this point I don't know how to begin with.