To show that $\mathbb{Z}[\zeta_5]$ is norm-Euclidean, you need to show that for every $x\in\mathbb{Q}(\zeta_5)$ then there is some $y\in\mathbb{Z}[\zeta_5]$ with $\vert N(x-y)\vert < 1$. Here $N(\cdot)$ is the standard norm (product of conjugates in $\mathbb{Q}(\zeta_5)$.
As you mention, it is made easier in $\mathbb{Q}(\sqrt{-1})$ because the algebraic integers form a lattice. In more general numbers fields, you can still construct a lattice by considering the set of all real and complex embeddings. In this case, there are 4 complex embeddings $\mathbb{Q}(\zeta_5)\mapsto\mathbb{C}$ corresponding to $\zeta_5\mapsto\zeta_5^r$ ($r=1,2,3,4$). Write these as two embeddings, $\sigma_r(\zeta_5)=\zeta_5^r$ ($r=1,2$), together with their two complex conjugates $\bar\sigma_r$. Then, you can define a map $ \begin{align} &f\colon\mathbb{Q}(\zeta_5)\to\mathbb{C}^2,\\ &f(x)=(\sigma_1(x),\sigma_2(x)). \end{align} $ It is clear that $f(x+y)=f(x)+f(y)$ ($f$ is linear as a map of vector spaces over $\mathbb{Q}$). In fact, the image of $\mathbb{Z}[\zeta_5]$ is a lattice. Also, defining $\theta\colon\mathbb{C}^2\to\mathbb{R}$ by $\theta(x,y)=\vert xy\vert^2$, you can write the norm in $\mathbb{Q}(\zeta_5)$ as $N(x)=\theta(f(x))$. The problem reduces to trying to prove that for every $x\in\mathbb{C}^2$ there is an element $y$ of the lattice $\Lambda\equiv f(\mathbb{Z}[\zeta_5])$ satisfying $\theta(x-y) < 1$.