Let $X$ be a elliptic curve over a field $k$ and $f$ be a holomorphic function on $X$, if $f$ only has two zeros $P$ and $Q$, then I'd like to know how to prove see that $P$ is a torsion point of $X$.
This question is motivated by the answer of Peter McNamara in this question.
Your statement isn't quite correct - let me show the linked claim, which is for a rational function $f$ on an elliptic curve $X$, the divisor of $f$ being supported only at $P$ and $O$ implies that $P$ is torsion.
The key is that if $A$ and $B$ are two points on $X$ with $A+B=C$ in the group law, then the divisors $A+B$ and $C+O$ are linearly equivalent: take the ratio of the functions cutting out the line through $A$ and $B$ which has third intersection point $-C$ with $X$, and the line through $O$ and $C$, which has third intersection point $-C$ with $X$ to get a rational function demonstrating this linear equivalence.
Next, a line bundle $\mathcal{O}_X(D-E)$ for effective divisors $D=\sum_{i=1}^n D_i$ and $E=\sum_{i=1}^n E_i$ of equal degree has a nonzero global section section iff it is trivial, which occurs iff the sum of the points in $D$ under the group law equals the sum of the points in $E$ under the group law, since $$\mathcal{O}_X(D-E)\cong \mathcal{O}_X([\sum D_i]+(n-1)O - [\sum E_i]-(n-1)O)$$ by repeated applications of the previous claim.
In our case, the line bundle of which $f$ is a section is one of the form $\mathcal{O}_X(nO-nP)$: $f$ has a pole of some order at $O$ and a zero of the same order at $P$, and no other poles or zeroes, so these must be of the same order, since the divisor of a rational function on a curve is of degree zero. But by the above, this says that $[nO]=[nP]$, which is the same as saying that $[nP]=O$, or that $P$ is $n$-torsion.