Let $p_n$ denote the probability of returning to the origin after n steps. If n is odd, $p_n = 0$.
The main insight is that $\sum_{n=0}^{\infty}p_{2n}$ is asymptotically ~ $C \cdot \frac{1}{n^{d/2}}$ for some constant C (this may be shown using Stirling's formula). This converges if and only if n >= 3.
Assuming this, is the following proof correct? It is simpler than the other proofs I know.
Denote by Q the probability that the walk never returns to the origin, and by $P_F$ the probability that the walk returns to the origin only finitely many times. Then:
$P_F = \sum_{n=0}^{\infty}P((x_{2n} = 0) \wedge (x_i \neq 0 \text{ for all } i > 2n)) = \sum_{n=0}^{\infty}P(x_{2n} = 0)\cdot P(x_i \neq 0 \text{ for all } i > 2n | x_{2n} = 0) = \sum_{n=0}^{\infty}P(x_{2n} = 0)\cdot P(x_i \neq 0 \text{ for all } i > 0 | x_0 = 0) = \sum_{n=0}^{\infty}p_{2n}\cdot Q = Q\cdot \sum_{n=0}^{\infty}p_{2n}$
If d<=2, $\sum_{n=0}^{\infty}p_{2n} = \infty$. Since $0\leq P_F \leq 1$, we must have Q=0 (and therefore also $P_F = 0$).
On the other hand, if d >= 3, $\sum_{n=0}^{\infty}p_{2n} $ converges. By Borel-Canteli, this implies that $P_F = 1$, and therefore $ Q = \frac{1}{\sum_{n=0}^{\infty}p_{2n}} > 0$.