How to describe the embedding? Start with the interval $[0, 2]$ and glue the points $0, 1$ and $2$. Describe the
equivalence relation $\mathcal{R}$ encoding this gluing and let $X = [0, 2]/\mathcal{R}$. Describe an embedding of $X$ in $\mathbb R^2$.
Equivalence relation is
\begin{align*}
    x \mathcal{R}y \Leftrightarrow \begin{cases} x=y \\
    x, y \in \{0, 1, 2\}.
    \end{cases}
\end{align*}
It is easy to see that $X$ is homeomorphic to the double circle. But how to describe the embedding?
 A: In the comments, you've written out a formula for a quotient map $f$ from $[0,2]$ to the double circle, whose point pre-images are identical to the equivalence classes of $\mathcal R$, i.e. for all $s,t \in [0,2]$ we have $s \mathcal R t$ if and only if $f(s)=f(t)$ (and $f$ is a quotient map, meaning, a surjective map such that a subset $A \subset $ (double circle) is open in the double circle if and only if $f^{-1}(A) \subset [0,2]$ is open in $[0,2]$).
It follows that the quotient space $X = [0,1] / \mathcal R$ is homeomorphic to the double circle. In fact even more follows: the formula
$$F[x]=f(x)
$$
is a well-defined formula for a homeomorphism
$$F : X \to \text{(double circle)}
$$
where the input to $F$ is an equivalence class $[x]$ represented by a point $x \in [0,2]$. These facts are all immediate consequences of the universality theorem for quotient maps.
A: The map $f : [0,2] \to \mathbb R^2 = \mathbb C, f(t) = - 1 + e^{2\pi i t}$ for $t \in [0,1]$, $f(t) = 1 + e^{2\pi i t}$ for $t \in [1,2]$ (which is the same as in your comment) is continuous and its image $D = f([0,2])$ is an embedded double circle.  Since $[0,2]$ is compact and $D$ is Hausdorff the map $q : [0,2] \stackrel{f}{\to} D$ is a closed map, therefore a quotient map. It is easy to see that $q(x) = q(y)$ iff $x \mathcal R y$. This means that there exists a unique homeomorphims $h : [0,2]/\mathcal R \to D$ such that $h \circ p = q$. See for example When is a space homeomorphic to a quotient space?
