You can certainly embed the real line, with perpendicular circles at the integer points, into $\mathbb{R}^3$, and this is homotopy equivalent to the wedge you want.

Let $W$ be the expanding `inverse Hawaiian earring'.

Then the obvious map $\bigvee S^1\to W$
is a homotopy equivalence. To define the inverse,
let $Z$ be the intersection of $W$ with
a small closed rectangle around the origin;

collapse $Z$ to a point, then map the resulting quotient back to
$\bigvee S^1$ by the inverse of the obvious map.

We have to check that $W\to \bigvee S^1$ is continuous.

If $U\subseteq \bigvee S^1$ is open
and does not contain the basepoint $\star$,
then the preimage is obviously open.
If $\star\in U$, then the preimage is the
preimage of $U- \star$ (which is open) together with
a neighborhood of $Z$, which is also open.

Since $Z$ is contractible and its
inclusion into $W$ is a cofibration,
the composite $W \to \bigvee S^1\to W$
is homotopic to $\mathrm{id}_{W}$.
In essentially the same way, the composite
$\bigvee S^1\to W\to \bigvee S^1$
is homotopic to the identity.