The questions is
Show that if $X$ is compact and all fixed points of $X$ are Lefschetz, then $f$ has only finitely many fixed points.
n.b. Let $f: X \rightarrow X$. We say $x$ is a fixed point of $f$ if $f(x) = x$. If $1$ is not an eigenvalue of $df_x: TX_x \rightarrow TX_x$, we say $x$ is a Lefschetz fixed point.
I have proved that $x$ is a Lefschetz fixed point of $f$ if and only if $\mathrm{graph}(f)$ and $\Delta = \mathrm{graph}(\mathrm{identity})$ intersect transversally at $(x,x) \in X \times X$, but not sure how to proceed.
Thank you!
Some key points for this answer is contributed by @tessellation:
Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can think of $f$ as a map from open ball in $\mathbb{R}^n,\,\,$ (say $B$) to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.
If $g$ is a diffeomorphism then $g$ is bijective in that small neighborhood. If $f$ has another fixed point a in that neighborhood then $g(a)=0$. This will contradict the bijcetivity of $g$. Therefore the Lefschetz fixed point is a 0-manifold. In other words, isolated.
Then we consider such an open cover of $X$: the open set around each fixed points, and the complement of the set of fixed poinots. Since $X$ is a compact manifold, so there must be finitely many of subcovers. Therefore, there are finitely many Lefschetz fixed points.