If $\operatorname{Hom}(\delta_V, \delta_W) = 0$, then $\mathfrak{C}(\delta_V) \cap \mathfrak{C}(\delta_W) = 0.$ Let $(A, \Delta)$ be a Hopf $^*$-algebra and $\delta_V: V \to V \otimes A$ and $\delta_W: W \to W \otimes A$ be two corepresentations of $(A, \Delta).$
Assume that the space of intertwiners $\operatorname{Hom}(\delta_V, \delta_W) = 0$, then is it true that $\mathfrak{C}(\delta_V) \cap \mathfrak{C}(\delta_W) = 0$, where
$$\mathfrak{C}(\delta) = \operatorname{span}\{(f \otimes\operatorname{id}_A)(\delta(v)): f \in V', v \in V\}$$
denotes the space of matrix coefficients of the corepresentation $(V, \delta).$
 A: As per e.g. An Algebraic Framework for Group Duality, A. Van Daele:

The Haar state $h:A\rightarrow \mathbb{C}$ is faithful in the sense that for $f\in A$,
$$h(f^*f)=0\implies f=0.$$

At least in the case where $\delta_V$ and $\delta_W$ are  irreducible and unitary, we can simply take non-zero $f\in\mathfrak{C}(\delta_V)\cap\mathfrak{C}(\delta_W)$ and calculate, using Timmermann 3.2.6:
$$h(f^*f)=0,$$
viewing the first $f\in \mathfrak{C}(\delta_V)$ and the second in $\mathfrak{C}(\delta_W)$. It follows that $f=0$.
I am not sure about the non-unitary case. Perhaps Timmermann explains something here (but I am away from my references).
A: As mentioned, the present question is related to Question on characterising CQGs. Since the questions are related, the answers are also.
I will work in the setting of Theorem 3.2.12 from Timmerman's book "An invitation to quantum groups and duality".

Let $(A,\Delta)$ be the given Hopf $*$-algebra.
As in (v) in Theorem 3.2.12, let $(\delta_\alpha)_{\alpha\in I}$ us fix an exhaustive set of irreducible unitary corepresentations
$\delta_\alpha:U_\alpha\to U_\alpha\otimes A$, which are pairwise inequivalent. They are indexed by some set $I$.
(As in Proposition 3.2.3 in loc. cit. we have a suitable inner product on $U_\alpha$ making $\delta_\alpha$ unitary.)
Let us use alternative notations
$C(V)=\mathfrak C(\delta_V)$,
$C(W)=\mathfrak C(\delta_W)$.
Since
$\Delta C(V)\subseteq C(V)\otimes C(V)$,
$\Delta C(W)\subseteq C(W)\otimes C(W)$,
we obtain restricted corepresentations of $\Delta:A\to A\otimes A$:
$$
\begin{aligned}
\Delta\Big|_{C(V)} &: C(V)\to C(V)\otimes A\ ,
\\
\Delta\Big|_{C(W)} &: C(W)\to C(W)\otimes A\ ,
\ .
\end{aligned}
$$
Let now $V_1\ne 0$ be a non-trivial minimal invariant subspace of $C(V)$.
So $\Delta\Big|_{V_1}:V_1\to V_1\otimes A$ is irreducible, thus isomorphic to some $\delta_{\alpha_1}$ for some $\alpha_1=\alpha(V_1)$ in the index list $I$.
Proposition 3.2.11 (mentioned in the other answer) shows then
$$
\Delta\Big|_{V_1}\cong\delta_{\alpha_1}^{\boxplus n(\alpha_1)}\ .
$$
Now i need finite dimensionality of $V$, or some Hermitian product (so that orthocomplements of invariant subspaces can be built, and are again invariant) in order to write $V$ as a direct sum of spaces $V_1,V_2,\dots$ - and in this case we have for instance
$$
\begin{aligned}
&\mathfrak C(\delta|_{V_1\oplus V_2})
\\
&=  
\operatorname{Span}\{\ 
(f \otimes\operatorname{id}_A)\delta(v)\ :\ 
f \in (V_1\oplus V_2)'\ ,\ v \in V_1\oplus V_2\ 
\}
\\
&=  
\operatorname{Span}\{\ 
((f_1\oplus f_2) \otimes\operatorname{id}_A)\delta(v_1\oplus v_2)\ :\ 
f_1\in V_1'\ , f_2\in V_2'\ ;\ v_1 \in V_1\ ,\ v_2\in V_2\ 
\}
\\
&=  
\operatorname{Span}\{\ 
(f_1\otimes\operatorname{id}_A)\delta(v_1)
+
(f_2\otimes\operatorname{id}_A)\delta(v_2)
\ :\ 
f_1\in V_1'\ , f_2\in V_2'\ ;\ v_1 \in V_1\ ,\ v_2\in V_2\ 
\}
\\
&=\mathfrak C(\delta|_{V_1}) + \mathfrak C(\delta|_{V_2})
\ .
\end{aligned}
$$
Some (transfinite) inductive argument should give then
$$
C(V)
=\mathfrak C(\delta)
=\mathfrak C(\delta|_V)
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta|_{V_1})
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha_1}^{\boxplus n(\alpha_1)})
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha_1})\ .
$$
(Equivalent corepresentations have the same matrix elements.)
A similar relation holds for $W$, so:
$$
\begin{aligned}
\mathfrak C(\delta_V)=
C(V) &= 
\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha(V_1)})
=\sum_{\alpha\in I(V)}
\mathfrak C(\delta_{\alpha})
=\bigoplus_{\alpha\in I(V)}
\mathfrak C(\delta_{\alpha})
\subseteq
\bigoplus_{\gamma\in I}
\mathfrak C(\delta_{\gamma})
\ ,
\\
\mathfrak C(\delta_W)=
C(W) &= \sum_{\substack{W_1\subseteq W\\W_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha(W_1)})
=\sum_{\beta\in I(W)}
\mathfrak C(\delta_{\beta})
=\bigoplus_{\beta\in I(W)}
\mathfrak C(\delta_{\beta})
\subseteq
\bigoplus_{\gamma\in I}
\mathfrak C(\delta_{\gamma})
\ ,
\end{aligned}
$$
where $I(V)$, $I(W)$ are subsets of the total index set $I$.
If there is now an element $u$ in the intersection of the above two spaces,
then there is a common $\gamma\in I(V)\cap I(W)$, we associate the corresponding minimal spaces $V_1\subseteq C(V)\subseteq A$, $W_1\subseteq C(W)\subseteq A$, and obtain a non-trivial intertwiner after writing
$\Delta|_{V_1}\cong\delta_\gamma^{\boxplus n(V_1)}$,
$\Delta|_{W_1}\cong\delta_\gamma^{\boxplus n(W_1)}$.
The projectors on the $\boxplus$-pieces are also intertwiners, so using kernels and orthocomplements we find a "same piece" $\delta_\gamma$ inside both $\Delta|_{V_1}$ and $\Delta|_{W_1}$.
Despite of the chosen notations, $V_1$ lives inside $C(V)\subseteq A$, not inside $V$. To connect with $V$, we use for some suitable $f\ne 0$ the map $T_f$ between corepresentations:
$\require{AMScd}$
\begin{CD}
@. V @>\delta_{V}>> V\otimes A\\
@. @V\delta_{V} VV @VV\delta_{V}\otimes \operatorname{id}_AV\\
@. V\otimes A @>\operatorname{id}_V\otimes\Delta>> V\otimes A\otimes A\\
@. @Vf\otimes \operatorname{id}_A VV @VV f\otimes \operatorname{id}_A\otimes \operatorname{id}_AV\\
@. \Bbb C\otimes A @>1\otimes\Delta>> \Bbb C\otimes A\otimes A\\
@. @| @|\\
u\in V_1\subseteq C(V)
@>\subseteq>> A @>\Delta>> A\otimes A\\
\end{CD}
Here, $T_f$ is the composition of the maps in the left vertical column. (It is mentioned in loc. cit. at the end of §3.1.1.) And $f$ is chosen so that $u$ is in the image of $T_f$. Using $T_f$ as a bridge, building kernel
and orhtocomplement, we manage get "a piece" $\delta_\gamma$ inside $V$ (i.e. inside $\delta_V$).
A similar argument realizes "a piece" $\delta_\gamma$ inside $W$.
Projectors onto $\boxplus$-pieces can be used to construct intertwiners, so we obtain an intertwiner between $\delta_V$ and $\delta_W$ using the identity of $\delta_\gamma$ realized as above "as a piece" inside them.
This contradicts the assumption $\operatorname{Hom}(\delta_V, \delta_W) = 0$.

As mentioned in the other answer, i am far away from home. The above argument is somehow K-theoretical, splits as much as possible the linear algebra into the corepresentative atoms $(\delta_\alpha)_{\alpha\in I}$, and builds bridges to them. I saw no direct path (from $V$ to $W$) for a solution. Hope that the arguments are OK.
