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.

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.