Inverse limit of sequence of fibrations

To be honest, I have trouble finding out what is the needed argument, but I'll try to give another one. I will consistently use radial coordinates on the disk, ie. $(r, \phi) \in D^{k}$, where $r$ is the distance from the centre and $\phi \in S^{k-1}$.

Let $\tilde{F}_{n-1}: S^{i} \times I \rightarrow X_{n-1}$ given by $\tilde{F}_{n-1}(\phi, t) = F_{n-1}(t-1, \phi)$ be the contracting homotopy. By the fibration property there exists a lift of this homotopy to $X_{n}$, say $G_{n}$, such that $G_{n} |_{S^{i} \times \{ 0 \}} = f_{n}$.

Let $g_{n} = G_{n} |_{S^{i} \times \{ 1 \}}$. Since $\tilde{F}_{n-1} |_{S^{i} \times \{ 1 \}}$ maps to the basepoint, we see that the image of $g_{n}$ is contained in the fibre $F \subseteq X_{n}$. The map $\pi_{i+1}(X_{n}) \rightarrow \pi_{i+1}(X_{n-1})$ is surjective and hence by the long exact sequence of homotopy the map $\pi_{i}(F) \rightarrow \pi_{i}(X_{n})$ is injective. Since $g_{n}$ is null in $\pi_{i}(X_{n})$ (being homotopic to $f_{n}$), it must be also null in $F$ and thus there is a contracting homotopy $H_{n}: S^{i} \times I \rightarrow F$.

Now define a map $F ^\prime _{n}: D^{i+1} \rightarrow X_{n}$ by

$F ^\prime _{n}(r, \phi) = G_{n}(\phi, 2(1-r))$ for $r \geq \frac{1}{2}$

$F ^\prime _{n}(r, \phi) = H_{n}(\phi, 2(\frac{1}{2}-r))$ for $r \leq \frac{1}{2}$,

After a moment of thought, we see that $p_{n} F^\prime _{n}$ is homotopic to $F_{n-1}$. Using the homotopy lifting for $(D^{i+1}, S^{i})$ we may deform it $rel \ S^{i}$ to a map $F_{n}$ such that $p_{n} F_{n} = F_{n-1}$, which is what we wanted to do.