I know that this is a well known result, but where can I find a proof? I am also interested to see more general non-embedding results of this type.
Theorem. Let $Y$ be the union of two segments with one segment being attached in the middle of the other one. Then there is no topological embedding of the space $Y\times Y$ into $\mathbb{R}^3$.
It was a homework problem in my undergraduate (sophomore) topology class taught by Professor Karol Sieklucki at University of Warsaw!
If it did embed as $Y \times Y \subset \mathbb{R}^3$, then you could choose a sufficiently small sphere $S$ in $\mathbb{R}^3$ centered at the point $(y_0,y_0)$, where $y_0$ is the point in where the two segments are glued. Then $S \cap Y \times Y \subset S$ is an embedding of the complete bipartite graph $K_{3,3}$ into a two sphere. But this graph is not planar, which is a contradiction. This argument seems to assume the embedding is sufficiently nice, e.g. simplicial with respect to some triangulations of $Y \times Y$ and $\mathbb{R}^3$. Maybe others can share a different argument that is valid for any topological embedding.
Theorem 10 of "An Alternative Proof that 3-Manifolds Can be Triangulated" by R.H. Bing [Ann. of Math. Vol. 69, (1959), pp. 37-65] states that any topologically embedded simplicial 2-complex $K$ in a triangulated 3-manifold $M$ is near a PL embedding of $K$ into $M$.
In particular, a $2$-dimensional finite simplicial complex has a topological embedding in $\mathbb R^3$ if and only if it has a PL embedding in $\mathbb R^3$. Together with Robert Bell's answer this resolves the question.
I will show that $Y^k$ cannot embed in $\mathbb{R}^p$ when $p < 2k$.
Write $\mathrm{Conf}(2,X) = \{(x,x') \in X^2 \, | \; x \neq x'\}$ for the deleted diagonal of $X$, and note that any injection $Y \hookrightarrow X$ induces an $S_2$-equivariant map $\mathrm{Conf}(2,Y) \to \mathrm{Conf}(2, X)$.
We need the following corollary of Theorem 1.8 from my paper "Configuration space in a product".
If an embedding $A \subseteq B$ induces a homotopy equivalence of pairs $(A^2, \mathrm{Conf}(2,A)) \simeq (B^2, \mathrm{Conf}(2,B))$, then it also induces a homotopy equivalence $\mathrm{Conf}(2,A^k) \simeq \mathrm{Conf}(2,B^k)$ for all $k \geq 0$.
Here is a short proof of the corollary in the case $k=2$.
Two points in $A^2$ are distinct if and only if they are distinct in their first coordinate, or their second coordinate, or both. In other words, $\mathrm{Conf}(2,A^2)$ is covered by the two open subsets $A \times \mathrm{Conf}(2,A)$ and $\mathrm{Conf}(2,A) \times A$, and the intersection of these sets is $\mathrm{Conf}(2,A) \times \mathrm{Conf}(2,A)$. As a consequence, $\mathrm{Conf}(2,A^2)$ is the homotopy pushout of a diagram that depends only on the pair $(A^2, \mathrm{Conf}(2,A))$. By our assumption, the inclusion $A \subseteq B$ induces a pointwise homotopy equivalence on these pushout diagrams, and hence on homotopy pushouts. This concludes the proof for $k=2$; the case of general $k$ requires a larger homotopy colimit, but is otherwise similar.
"Reordering the cars in the driveway twice" gives an equivalence $\mathrm{Conf}(2,Y) \simeq \mathrm{Conf}(2, \mathbb{R}^2)$, (even when restricted to the driveway and a little part of the street, two cars may wind around each other), leading to an equivalence of pairs $$ (Y^2, \mathrm{Conf}(2,Y)) \simeq (\mathbb{R}^4, \mathrm{Conf}(2,\mathbb{R}^2)) $$ induced by the usual inclusion $Y \subset \mathbb{R}^2$. By the corollary, $\mathrm{Conf}(2, Y^k)$ is homotopy equivalent to $\mathrm{Conf}(2, \mathbb{R}^{2k})$, and moreover, this map is $S_2$ equivariant.
Since $\mathrm{Conf}(2, \mathbb{R}^{p}) \simeq_{S_2} (S^{p-1}, \tau)$, where $\tau$ denotes the antipodal action, and similarly $\mathrm{Conf}(2,Y) \simeq_{S_2} \mathrm{Conf}(2, \mathbb{R}^{2k}) \simeq_{S_2} (S^{2k-1}, \tau)$, any embedding $Y^{k} \subseteq \mathbb{R}^p$ induces an $S_2$-map $$ (S^{2k-1}, \tau) \to (S^{p-1}, \tau), $$ which is impossible for $p<2k$ by the Borsuk-Ulam theorem.
