Infinite graph $G=(\omega, E)$ with $\delta(G) = \aleph_0$ and no perfect matching

Question

Is there an infinite graph $G=(\omega, E)$ on the vertex set $\omega$ such that every vertex has infinite degree, but $G$ has no perfect matching?

Answer 1

I suspect the answer is no. Here's a proof attempt, which I confess I'm not fully confident on:

We will construct a matching that contains every vertex. Let $V_0 = \mathbb{N}$.

We build our matching $M=\emptyset$ recursively. Let $u_1 = 1$, and let $v_1$ be the smallest number adjacent to $u_1$, and add the edge $u_1v_1$ to $M$. Now let $V_1 = V_0 - \{u_1,v_1\}$, and define $u_2$ to be the smallest natural number in $V_1$ (so $u_1$ will be either 2 or 3), and let $v_2$ be the smallest number in $V_1$ adjacent to $u_2$. Add to $M$ the edge $u_2v_2$, and let $V_2 = V_1 - \{u_2, v_2\}$. Repeat this process for all $n\in\mathbb{N}$ to construct the set $M\subset E$.

The set $M$ of edges obtained in this way must be a matching, and the vertex $k$ is always incident to one of the first $k$ edges added to $M$, so $M$ is perfect.

EDIT: Thank-you to bof for pointing out an error in the initial proof, and providing a much simpler approach!