The Hawaiian earring $\mathbf{H}$ is the union of the circles $C_n \subseteq \mathbb{R}^2$ with radius $\frac{1}{n}$ and centre $(\frac{1}{n}, 0)$. The Griffiths twin cone is then $\mathbf{TC} := C\mathbf{H} \vee C\mathbf{H}$, where $C \mathbf{H}$ is the cone of $\mathbf{H}$ (i.e. the quotient of $\mathbf{H} \times I$ by identifying $\mathbf{H} \times 1$ to a point). The wedge point is taken to be the point corresponding to $(0,0,0)$ in each copy of $C\mathbf{H}$. Griffiths proved in his 1954 paper The fundamental group of two spaces with a common point that $\mathbf{TC}$ is not simply connected.
I am interested in finding an elementary proof of this result (preferably without using van Kampen's theorem, but not necessarily). In the first and second copy of $C\mathbf{H}$ in the wedge, let $C_n$ and $C_n'$ (by an abuse of notation) be the circles corresponding to the original circle $C_n$ in $\mathbf{H}$. Then let $\alpha \colon I \rightarrow \mathbf{TC}$ be a loop traversing $C_1$ during $[0, \frac{1}{2}]$, $C_1'$ during $[\frac{1}{2}, \frac{2}{3}]$, $C_2$ during $[\frac{2}{3}, \frac{3}{4}]$, $C_2'$ during $[\frac{3}{4}, \frac{4}{5}]$, and so on. I know that this loop is not nullhomotopic, but I would like to have an elementary proof that any homotopy of $\alpha$ cannot be a nullhomotopy.
Intuitively it seems clear to me that this is true because the loop would have to jump to the different cone points infinitely often, which should violate continuity, but I really don't know how to show this. The space is compact, so one can't use arguments similar to the ones used to show that the harmonic archipelago is not simply connected.