For a group $G$ we define $Z_1(G) = Z(G)$ and then
$$Z_{i+1}(G) = \pi^{-1}(Z(G/Z_i(G)))$$
with that we can build a normal series
$$\{1\} \triangleleft Z_1(G) \triangleleft\cdots \triangleleft Z_n(G) \triangleleft \cdots $$
and I am asked to prove that $Z_i(G) \triangleleft G\ \forall i$.
I thought this would be done by induction and so I started by stating that for $i = 1$ it comes directly by definition.
Then I proceeded to show that $Z_i(G) \triangleleft G \implies Z_{i+1}(G) \triangleleft G$:
First note that $z \in Z_{i+1}(G) \iff \pi(z) \in Z(G/Z_i(G))$ that is, $\pi(z)$ commutes with every element in $G/Z_i(G)$. Having said that, let $z \in Z_{i+1}(G), g \in G$:
$$\pi(gzg^{-1}) = \pi(g)\pi(z)\pi(g)^{-1} = \pi(z)\pi(g)\pi(g)^{-1} = \pi(z)$$
because $\pi(z)$ commutes with every element; if $\pi(gzg^{-1}) = \pi(z)$ then $\pi(gzg^{-1})$ also commutes with every element, hence $gzg^{-1} \in Z_{i+1}(G)$ and so $Z_{i+1}(G) \triangleleft G$.
It feels as if I didn't use the induction hypothesis, but my proof also seems correct, so is it the case where a direct proof was possible? i.e. I could prove $Z_i(G) \triangleleft G$ directly for all $i$? Or did I use the induction hypothesis without even realizing?
You used that $Z_i(G)$ is normal when you wrote $G/Z_i(G)$. Other than that, this result is true for any normal subgroup i.e., if $N$ is normal in $G$ then $H=\pi^{-1}(Z(G/N))$ is also normal in $G$.