FALSE: every compact path-connected subset of ${\mathbb{R}}^2$ has a universal cover

By [F1], $\mathbb{H}$ satisfies the compactness and path-connectedness hypotheses of the proposed statement.

Let $U$ be any neighbourhood of $0$ in $\mathbb{H}$. By [F4], there is an ambient open set $O\subseteq {\mathbb{R}}^2$ with $0\in O$ and $O\cap \mathbb{H}\subseteq U$. By [F5], choose $\epsilon >0$ with $B(0,\epsilon )\subseteq O$, so $\mathbb{H}\cap B(0,\epsilon )\subseteq U$. Since $C_n\subseteq {\bar{B}}_2(0,2\mathrm{/}n)$, [L4] gives an $n\ge 1$ with $C_n\subseteq U$.

Let $j:C_n\hookrightarrow \mathbb{H}$ be inclusion and $r_n:\mathbb{H}\to C_n$ the retraction of [F2]. Functoriality gives $(r_n{)}_{\ast }\circ j_{\ast }=\mathrm{id}$, so $j_{\ast }$ is injective. The pointed affine homeomorphism $z\mapsto (1\mathrm{/}n,0)-(1\mathrm{/}n)z$ carries the geometric unit circle based at $(1,0)$ onto $C_n$ based at $0$ and carries the once-around loop of [L2] to a loop ${\mathrm{\ell }}_n$ in $C_n$. By [L2] and [L3], $[{\mathrm{\ell }}_n]\mathrm{\ne }1$ in ${\pi }_1(C_n,0)$; injectivity of $j_{\ast }$ therefore makes its image nontrivial in ${\pi }_1(\mathbb{H},0)$. Since $C_n\subseteq U$, the same ${\mathrm{\ell }}_n$ is a loop in $U$.

Since every neighbourhood $U$ of $0$ contains such a loop, no inclusion-induced map ${\pi }_1(U,0)\to {\pi }_1(\mathbb{H},0)$ is trivial. Thus $\mathbb{H}$ is not semilocally simply connected at $0$.

By [L1], the Hawaiian earring has no universal cover. Together with step 1.1, it is a compact path-connected planar counterexample to the proposed universal claim.