Here is a more naive solution, as least if the sequence is countable. Let
$\Bbb N \subset \Bbb R^2$ be the embedding defined by the sequence.
Then there is an isotopy from this embedding to the standard inclusion into
the $x$-axis (inductively move the points one by one through embeddings to each integer point
on the $x$-axis. The homotopy type of the complement does not change through an isotopy. Furthermore, the homotopy type of the complement of the set of integer points on the $x$-axis
clearly has the homotopy type of a wedge of circles (this can be seen, e.g., as follows: (1). the complement of the integer points in $\Bbb R$ has the homotopy type of an infinite wedge of zero-spheres, (2). passing from $\Bbb R^1$ to $\Bbb R^2$ has the effect of suspending the complement).

Addendum: maybe the following is a better way to see the answer. Let $X = \lbrace x_n \rbrace$
be the sequence.

We can find a sequence of spaces $D_1 \subset D_2 \subset \cdots $ exhausting $\Bbb R^2$ such that $D_k$ is homeomorphic to a closed disk, $D_k$ is embedded in the interior of
$D_{k+1}$ and
$X$ meets each $D_k$ in its interior. Let $X_k = X \cap D_k$, and let $C_k$ be its complement in $D_k$. Then $C_k$ is a finite wedge of circles up to homotopy, $C_k \subset C_{k+1}$ and and $C:= \cup_k C_k$ is the complement of $X$ in $\Bbb R^2$.

Furthermore, the inclusion $C_k \subset C_{k+1}$ is a cofibration and admits a retraction, so we can write $$C_{k+1} \simeq C_k \vee E_k$$ and $E_k$ is a finite wedge of circles.

Then $C$, which is a colimit of the $C_k$, coincides with the homotopy colimit of the $C_k$,
and with respect to the displayed identification we see that the homotopy colimit is a countable wedge of circles.