A metrizable example can be constructed as follows. In the plane consider the subset $$\Xi:=\big\{(x,\tfrac{2k+1}{2^n}):k,n\in\mathbb Z,\;x\in\mathbb R\setminus \tfrac1{2^n}\mathbb Z\big\}.$$

It is clear that $\Xi$ contain (countably many) topological copies of the real line.

There are at least two ways of proving that any non-empty open subset of $\Xi$ is homeomorphic to $\Xi$. One is more geometric and is due to Volodymyr Mykhaylyuk (from Chernivtsi). He observed that for any open set $U\subset \Xi$ and any connected component $C$ of $U$ the interval $C$ has a base of clopen neighborhoods homeomorphic to the strip $\Xi\cap(\mathbb R\times(-\sqrt{2},\sqrt{2}))$ in $\Xi$. Then $U$ can be decomposed into countably many pairwise homeomorphic clopen sets and the same can be done with the space $\Xi$.

Another way is more global. Just to prove a characterization theorem for the space $\Xi$:

**Theorem.** A topological space $X$ is homeomorphic to the space $\Xi$ if and only if 

1) $X$ is a metrizable space;

2) for any point $x\in X$ the connected component $C_x$ containing $x$ is homeomorphic to the real line;

3) the family $\mathcal C=\{C_x\}_{x\in X}$ of connected components of $X$ is countable;

4) for any point $x\in X$ and a neighborhood $O_x\subset X$ of $x$ there exists a non-empty set $V=\bigcup\{C\in \mathcal C:C\cap V\ne\emptyset\}$ in $O_x\setminus C_x$ such that $V$ is clopen in $X\setminus C_x$, $C_x\cup V$ is a neighborhood of $x$.

The proof of this characterization uses the back-and-forth argument: We enumerate the connected components and at the $n$-th step construct clopen neighborhoods of $n$-th intervals and establish the combinatorial correspondence between these neighborhoods. I will write down the details in a preprint (with many authors with whom I discussed this problem being on a Summer School in Carpathian mountains).

The characterization theorem implies the following corollary:

**Corollary.** Let $(C_n)_{n\in\omega}$ be a family of parwise disjoint curves in $\mathbb R^d$ such that each $C_n$ is homeomorphic to $\mathbb R$, is closed and nowhere dense in the union $X:=\bigcup_{n\in\omega}C_n$ and $\sum_{n=1}^\infty lenth(C_n)<\infty$. Then the space $X$ is homeomorphic to $\Xi$.