As a reference my definition of nilpotence is as follows.
Definition: A group $G$ is said to be nilpotent if it has a subnormal series $$ G = G_1 \trianglerighteq G_2 \trianglerighteq \dots \trianglerighteq G_n = \{1\} $$ for which $G_{i} \unlhd G$ and $G_{i-1}/G_i \subseteq Z(G/G_i)$ for all $1 \leq i \leq n$.
I've seen a few proofs on this site, but I don't see why at least part of this isn't just an application of the 3rd Isomorphism theorem.
Attempt: Firstly recall that by the correspondence isomorphism theorem it follows that every subgroup of $G/Z(G)$ has the form $H/Z(G)$ where $H$ is a subgroup of $G$ containing the centre. With this in mind, suppose that $G/Z(G)$ is nilpotent meaning it has a subnormal series $$ G/Z(G) = G_1/Z(G) \trianglerighteq G_2/Z(G) \trianglerighteq \dots \trianglerighteq G_n/Z(G) = \{1\} $$ satisfying $(G_i/Z(G)) \unlhd G/Z(G)$ and $(G_{i-1}/Z(G))/(G_i/Z(G)) \subseteq Z((G/Z(G))/(G_i/Z(G))$ for every $1 \leq i \leq n$. By the 3rd Iso. theorem we know that $$ (G_{i-1}/Z(G))/(G_i/Z(G)) \cong G_{i-1}/G_i \text{ and } (G/Z(G))/(G_i/Z(G)) \cong G/G_i, $$ and so the nilpotence of $G/Z(G)$ implies that $G_{i-1}/G_i \subseteq Z(G/G_i)$ and that $G_i \unlhd G$ for every $1 \leq i \leq n$. Moreover because $G_{i-1}/Z(G) \trianglerighteq G_i/Z(G)$ it follows that $G_i$ is a normal subgroup of $G_{i-1}$ and so we have a subnormal series for $G$ by $$ G = G_1 \trianglerighteq \dots \trianglerighteq G_n = Z(G) \trianglerighteq \{1\} $$ that witnesses the nilpotence of $G$.
Does this make sense? I'm not sure if the fact that $G_i/Z(G) \unlhd G/Z(G)$ implies $G_i \unlhd G$ is correct, I know that $G_i$ has to be a subgroup of $G$ containing the centre, but is it wrong to conclude it's normal? Thanks in advance for the help.