A proof of this property can be found in Proposition 2.2 of Huang's paper *Vertex operator algebras and the Verlinde conjecture*. This proof actually uses Zhu's algebras to simplify the discussions (similar to Zhu's proof of the linear independence of characters in *Modular invariance of characters of vertex operator algebras*) although I think this is not necessary. So let me give an elementary argument without using Zhu's algebras. Indeed I think this argument should also work for proving the injectivity of the sewings of (the vector spaces of) general conformal blocks.

Let me change your notations a little bit, and restate your claim as follows:

**Proposition**. Let $M_1,\dots,M_n$ be a list of inequivalent irreducible $V$-modules. For any $k=1,\dots,n$ we choose any intertwining operator $\mathcal Y_k$ of type $M_k\choose WM_k$. Assume that $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)q^{L_0})=0$ for any $w,z,q$. Then $\mathcal Y_k=0$ for all $k$.

*Proof*. It suffices to show that there exits a $\mathcal Y_k$ equaling $0$. Then, by induction, all $\mathcal Y_k$ are $0$.

For each $k$ we have a natural injective linear map $A_k:M_k\otimes\overline M_k\rightarrow End(M_k)$, where $\overline M_k$ is the contragredient (conjugate) $V$-module of $M_k$. Set $\mathfrak M=\bigoplus_kM_k\otimes\overline M_k$, and set $A=\bigoplus_k A_k$.
We then have $$A:\mathfrak M\hookrightarrow\bigoplus_k End(M_k).$$

Now let $\mathfrak N$ be the subspace of all $\xi\in\mathfrak M$ satifying that $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0$ for any $w,z$. (Here we regard the action of $\mathcal Y_{k_1}$ on $M_{k_2}$ as zero if $k_1\neq k_2$.)

$$\mathfrak N := \left\{\xi\in\mathfrak M:\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0\right\}$$
We claim that $\mathfrak N\neq0$. Indeed, since $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)q^{L_0})=0$, if we regard it as a series of $q$: $\sum_{\Delta}\sum_k Tr_{M_k(\Delta)}(\mathcal Y_k(w,z))q^\Delta$, then any coefficient $\sum_k Tr_{M_k(\Delta)}(\mathcal Y_k(w,z))$ must be $0$ 1. Choose a conformal weight $\Delta$ such that there exists $k$ such that $M_k(\Delta)$ is non-zero. Then we immediately have the non-zeroness of $\mathfrak N$.

Note that $\mathfrak M$ is indeed a $V\otimes V$-module. We now show that $\mathfrak N$ is a $V\otimes V$-submodule. Choose any $\xi\in\mathfrak N$. Then, for any $w,z$, since $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0$, we have $\sum_k Tr_{M_k}(\mathcal Y_k(Y(u,z_0-z)w,z)A(\xi))=0$ for any $u$ and any $z_0$ such that $0<|z_0-z|<|z|$. If moreover $0<|z_0-z|<|z_0|<|z|$, then $\mathcal Y_k(Y(u,z_0-z)w,z)=\mathcal Y_k(w,z)Y(u,z_0)$. Thus $$\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u,z_0)A(\xi))=0$$ for all $z_0$ satisfying $0<|z_0-z|<|z_0|<|z|$. By uniqueness of analytic continuations, we have $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u,z_0)A(\xi))=0$ for all $z_0$ satisfying only $0<|z_0|<|z|$. Take contour integrals of $z_0$. Then we obtain $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u)_mA(\xi))=0$ for any $m\in\mathbb Z$, where we write $Y(u,z_0)=\sum_{m}Y(u)_mz_0^{-m-1}$. Since $Y(u)_mA(\xi)=A((Y(u)_m\otimes 1)\xi)$, we've actually shown that $\mathfrak N$ is invariant under the action of $V\otimes \Omega$ (where $\Omega$ is the vacuum vector of $V$). Similarly $\mathfrak N$ is $\Omega\otimes V$-invariant. So $\mathfrak N$ is $V\otimes V$-invariant, i.e., a $V\otimes V$ submodule of $\mathfrak M$.

Since $\mathfrak M$ has irreducible decomposition $\mathfrak M=\bigoplus_k M_k\otimes\overline M_k$, and since $\mathfrak N$ is a non-trivial submodule of $\mathfrak M$, there exists $k$ such that $M_k\otimes\overline M_k\subset \mathfrak N$. Therefore, for any $w_1\in M_k,w'_2\in \overline M_k$, we have $\langle\mathcal Y_k(w,z)w_1,w'_2\rangle=Tr_{W_k}(\mathcal Y_k(w,z)A(w_1\otimes w_2'))=0$ for any $w,z$. Thus $\mathcal Y_k=0$.

*Q.E.D.*

1 We know that the coefficients $c_n$ of a convergent power series $f(z)=\sum_{n\in\mathbb N}c_nz^n$ are determined by the values of $f$. This is also true if $f$ can be written as a finite sum $f(z)=\sum_m z^{\mu_m}g_m(z)$ where each $g_m(z)$ is a convergent power series of $z$ and $\mu_m\in\mathbb C$. This is an easy exercise. See for example *A theory of tensor products for module categories for a vertex operator algebra, IV* Lemma 14.5 and section 15.4.