Every finite-dimensional Noetherian local ring has a system of parameters

If $d=0$, the local ring has no strict prime inclusion, so its maximal ideal $\mathfrak{m}$ is its only prime. Consequently $\sqrt{(0)}=\mathfrak{m}$, and the empty tuple generates $(0)$ and is a system of parameters by [L2].

Assume $d>0$ and that the theorem is known in dimensions $<d$. By [L1], choose $x\in \mathfrak{m}$ outside every minimal prime. Then $R\mathrm{/}(x)$ is a Noetherian local ring of some dimension $e\le d-1$. By the induction hypothesis, choose a system of parameters $({\bar{y}}_1,\dots ,{\bar{y}}_e)$ in $R\mathrm{/}(x)$ of length $e$. Lift these elements to $y_1,\dots ,y_e\in \mathfrak{m}$ and append $d-1-e$ zeros, obtaining a tuple $(x_2,\dots ,x_d)$ of length $d-1$. Padding does not change the ideal or its radical.

The ideal generated by $({\bar{x}}_2,\dots ,{\bar{x}}_d)$ has radical $\mathfrak{m}\mathrm{/}(x)$ in $R\mathrm{/}(x)$, so [L3] says the ideal $(x,x_2,\dots ,x_d)$ has radical $\mathfrak{m}$ in $R$. By [L2], the tuple $(x,x_2,\dots ,x_d)$ is a system of parameters.

Therefore every finite-dimensional Noetherian local ring has a system of parameters.