Let $X=[-1,1] \cup [-1,1] \cup D$ where the first segment lies on the x-axis, the second segment lies on the y-axis and $D$ is the disk of center (0,2) and radius 1.
Let $Y$ be a 1-dimensional cell complex. The goal is to prove that $X$ is not homeomorphic to an open subset $U$ of $Y$.
My try: This is how I am trying to do it. Suppose there exists $f$ a homeomorphism between $X$ and $Y$ then $X-{x}$ is also homeomorphic to $U-{f(x)}$. Hence it induces isomorphisms on homology groups and we have $H_k(U,U-{f(x)})= H_k(X,X-{x})$.
But $H_k(U,U-{f(x)})=H_k(Y,Y-{f(x)})$ by excision theorem and I feel like here we can use the fact that $Y$ is a 1-dim cell complex so we have $H_k(Y)=0$ for $k \ge 2$. I feel like something is missing. One idea might be to use the Long exact sequence with a specific point $x$. But I don't know if this is going to work.
Assume $X$ is homeomorphic to some open subset $U\subseteq Y$. Since $X$ contains an open $2$-dimensional disk $B$ (the interior of $D$) then we can restrict this homeomorphism to conclude that $B$ is homeomorphic to some open subset $V\subseteq U\subseteq Y$. And this can't be true by the dimension argument as we will see soon.
Now $Y$ is a cell complex of dimension $1$. Take the $0$-th skeleton $Y^0\subseteq Y$. Without the loss of generality, we may assume that none of the points in $Y^0$ are isolated in $Y$. That's because $B$ has no isolated points and homeomorphisms preserve isolated points. And so $V$ cannot have isolated points and thus we can just remove isolated points from $Y$ and only deal with the non-isolated part of the cell complex.
Therefore any open subset in $Y$ has to intersect nonemptly with $Y\backslash Y^0$. But $Y$ is of dimension $1$, meaning $Y\backslash Y^0$ is a (disjoint) union of $1$-dimensional open cells. Recall that a $1$-dimensional open cell is a homeomorphic image of $(0,1)$.
Therefore there is some open, $1$-dimensional cell $e\subseteq Y$ such that $e\cap V\neq\emptyset$. But any open subset of $(0,1)$ contains yet another copy of $(0,1)$ as an open subset. Meaning there is $e'\subseteq e\cap V$ such that $e'$ is open and $e'\simeq (0,1)$. And thus our homeomorphism $B\simeq V$ induces a homeomorphism $U'\simeq e'$ where $U'\subseteq B$ is an open subset of $B$, because $e'$ is an open subset of $V$. This cannot be true because $e'$ is of dimension $1$ while $U'$ is of dimension $2$ (more precisely: because of the invariance of domain).
Pretty much the same reasoning can be applied to any two cell complexes $X$ and $Y$ such that $\dim X>\dim Y$ to conclude that $X$ cannot be homeomorphic to an open subset of $Y$. Possibly to any subset, but I couldn't prove such statement.