let X =Hawaiian earring, consider CX be the cone with base X. Now take two coies of this space named as CZ and CY and z be the point where all circles of Z are tangent similarly for y.Now consider K= CZ V CY (wedge sum of two spaces identifying the point z with y). Now K is connected and observe that both CZ and CY are contractable but the fundamental group of K is not trivial.(a closed path oscillating back to forth from Y to Z around the decreasing circles is not null homotopic. You can get the reference form the paper- "THE COMBINATORIAL STRUCTURE OF THE HAWAIIAN EARRING GROUP" - bY J.W.CANNON & G.R.CONNER )

So now if you consider the inclusion map from CZ or CY to K then for give any covering space of K you can lift CZ or CY (by map lifting theorem). so it follows that any connected covering space of K is homemorphic to K. so K is its own universal cover. and fundamental of K is not trivial.

let $X$=Hawaiian earring, consider $CX$ be the cone with base X. Now take two copies of this space named as $CZ$ and $CY$ and $z$ be the point where all circles of $Z$ are tangent similarly for $y$. Now consider $K= CZ \wedge CY$ (wedge sum of two spaces identifying the point $z$ with $y$). Now $K$ is connected and observe that both $CZ$ and $CY$ are contractable, but $\pi_1(K)$ is not trivial. For example, a closed path oscillating back to forth from $Y$ to $Z$ around the decreasing circles is not null homotopic. A good reference for this is J.W.Cannon and G.R. Conner's paper- "The combinational structure of the hawaiian earing group."

Consider the inclusion map from $CZ$ or $CY$ to $K$. For any covering space $\tilde{K}$ of $K$, there exists a lift $CZ$ or $CY$ to $\tilde{K}$ by map lifting theorem. It follows that any connected covering space of $K$ is homeomorphic to $K$, and so $K$ is its own universal cover. However, as mentioned above $\pi_1(K)$ is not trivial.