Hawaiian earring is not homotopy equivalent to cone over subspace of $[0,1]$ Let $X = [0, 1]$, let $A = \{0\} \cup \{1/n : n \in \mathbb{N}\}$. Let $CA$ denote the cone on $A$. The spaces $X/A$ (the Hawaiian earring) and $X \cup CA$ are not homotopy equivalent; one way to see this is that their first homology groups are different.
$$
H_1(X\cup CA) = \widetilde {H}_1(X, A) \preceq \bigoplus_1^\infty \mathbb{Z}$$
where the second inclusion comes from considering the long exact sequence of homology of the pair $(X, A)$. On the other hand $H_1(X/A)$ is well-known to be uncountable and contains $\prod_1^\infty \mathbb{Z}$ as a subgroup.
Is there an intuitive purely topological reason why the two spaces are not homotopy equivalent?
 A: The fundamental group actually has a reasonable topology that makes it into a topological space. You topologize it as the quotient of the based mapping space by the equivalence relation generated by paths (AKA homotopies of loops).
For CW complexes, this topology will always be discrete, but for other spaces, like the ones you give, they do not have to be. Since you aren’t asking about pointed homotopy equivalence, let’s use the analogous unpointed fundamental set of homotopy classes of loops.
In the Hawaiian earring, there is a sequence of essential loops that converges to a constant loop, but in your other space it is clear that if we metrize it in a reasonable way, every essential loop has to be relatively long because it must wrap around a triangle with fixed height.
This argument can be formalized to show that no sequence of essential loops converges to the constant loop in the space you describe. Since any map of path connected spaces necessarily preserves the homotopy class of the constant loop, we know that any map between your two spaces fails to be a homotopy equivalence, since if it was it would induce a pointed homeomorphism between the fundamental sets. However, we just showed the basepoints were topologically different.
A: Here's an intuitive argument that their fundamental groups aren't the same; Connor Malin's answer is more rigorous.  Both spaces have infinitely many loops, but in $X/A$ any neighborhood of the special point contains all but finitely many of the loops, while in $X \cup CA$ the loops are all "large".  So in the former it's possible to continuously traverse all the loops consecutively in any order (for example, spending $1/2^n$ units of time traversing the $n$th loop).  I claim this is not the case in $X \cup CA$.  (Note that this doesn't really prove $X/A \not\simeq X \cup CA$, since it requires us to assume that the $n$th loop in $X/A$ corresponds to the $n$th loop in $X \cup CA$, and also to pretend that infinite products make sense in $\pi_1$.  The latter issue can be solved by topologizing the fundamental group as in Connor's answer.)
Although the loops in $X \cup CA$ share edges, they can be ordered in such a way that we must return to the special point between each pair. (For example, never do adjacent loops consecutively.)  Then suppose we have a path that traverses all the loops in such an order, choose times $t_1 < t_2 < \cdots$ when we're at the special point between loops, and let $t$ be the limit of a convergent subsequence. In any interval around $t$, we must both stay near the special point (by continuity) and traverse infinitely many loops (by construction), which is a contradiction.
