In this post, D. J. Obata gave an answer in which he says that it is a good exercise to prove that the constructed splitting $T_{\Lambda}M=\widetilde{E}^{u}\oplus\widetilde{E}^{s}$ is indeed a hyperbolic splitting.
So I attempted to prove this result, but I did not succeed and got stuck almost immediately.
Can someone elaborate on the answer of D. J. Obata? Many thanks in advance!
EDIT: I found a proof of a similar statement (see Lemma 2.3). Can someone explain the step: "By continuity of $Df$ there is a neighbourhood $U$ of $p$ and $V$ of $q$ such that $\|Df(f^{n}(z))|_{E_{f^{n}(z)}^{s}}\|<\lambda$ and $\|Df^{-1}(f^{n}(z))|_{E_{f^{n}(z)}^{u}}\|<\lambda$ for $f^{n}(z)\in U\cup V$."
You only need to note (in your case) that $f^n(q)$ approaches $p$ when $n\to+\infty$ and when $n\to-\infty$.
This lets you use the lambda lemma, which basically says that under iteration curves get close in the $C^1$ topology to the stable and unstable manifold of the hyperbolic fixed point. Thus the spaces that are going to be the stable and unstable spaces vary continuously in the $C^0$ topology and thus why "by continuity" in the proof of that Lemma 2.3. The case in Lemma 2.3 is lightly more general but nothing really changes.
In other words, what really matters is the lambda lemma, not that the derivative is continuous (the first is rather delicate while the second is simply an hypothesis).
Summing up, and answering to the question in your title: because of the lambda lemma. The rest is only the $C^1$ property (of the manifolds in the lambda lemma and of $f$).