Kernel of intertwiner is invariant (compact quantum groups) Before asking my question, let me introduce the relevant terminology.
Throughout, let $(A, \Delta)$ be a compact quantum group.
Definition: A representation $v$ on the Hilbert space $H$ is an element $v\in M(B_0(H)\otimes A)$ such that $(\text{id}\otimes \Delta)(v) = v_{(12)}v_{(13)}$. Here the subscripts with the brackets denote the leg numbering notation.
Definition: An intertwiner from the representation $(H_1, v_1)$ to the representation $(H_2, v_2)$ is an element in $B(H_1,H_2)$ such that $(x\otimes 1)v_1 =v_2(x \otimes 1).$
Definition: A closed subspace $K$ of $H$ is called invariant under the representation $(H,v)$ if $(e\otimes  1)v(e\otimes 1) = v(e\otimes 1)$ where $e$ is the orthogonal projection of $H$ onto $K$.
Question: Let $x: H_1 \to H_2$ be an intertwiner as above. Why Is $\ker(x)$ an invariant subspace of $H_1$?
 A: Let $e$ be the orthogonal projection onto $\ker(x)$.
If the result is not true, then there is $\xi\otimes\eta \in H_1\otimes K$ with
$$ (e\otimes 1) v_1 (e\xi\otimes\eta) \ne v_1 (e\xi\otimes\eta), $$
because the linear span of such vectors in dense in $H_1\otimes K$.  Here $K$ is some auxiliary Hilbert space such that we can regard $A\subseteq\mathcal B(K)$.  Similarly, there is $\xi'\otimes\eta'\in H_1\otimes K$ with
$$ \bigl((e\otimes 1) v_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr) \not=
\bigl( v_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr). $$
I write $(\cdot\mid\cdot)$ for the inner-product.
The map $T:\mathbb C \rightarrow K$; $\alpha\mapsto\alpha\eta'$ is bounded, and so there is an adjoint $T^*:K\rightarrow\mathbb C$ which is simply $K \ni \zeta \mapsto (\zeta\mid\eta')$.  Then $1\otimes T^*:H_1\otimes K \rightarrow H_1\otimes\mathbb C \cong H_1$ is bounded.  Consider
$$ \xi'' = (1\otimes T^*)v_1(e\xi\otimes\eta) \in H_1. $$
For any $\xi_0\in H_1$ we have that $(\xi''\mid\xi_0) = (v_1(e\xi\otimes\eta)\mid\xi_0\otimes\eta')$ and hence in particular
$$ (\xi''\mid\xi') \ne (\xi''\mid e\xi'). $$
Notice that $(x\otimes 1)v_1(e\otimes 1) = v_2(xe\otimes 1)=0$ and so
$(x\otimes 1)v_1(e\xi\otimes\eta) = 0$.  Hence
$$ 0 = \bigl((x\otimes 1)v_1(e\xi\otimes\eta)\bigm|\xi_0\otimes\eta'\bigr)
= \bigl(v_1(e\xi\otimes\eta)\bigm|x^*\xi_0\otimes\eta'\bigr)
= (\xi''\mid x^*\xi_0) = (x\xi''\mid\xi_0), $$
for any $\xi_0\in H_1$.  This shows that $x\xi''=0$ so $\xi''\in\ker(x)$ so $e\xi''=\xi''$, which gives the required contradiction.
A: While the answer of @Matthew Daws is certainly a good one, I want to give another perspective.

The classical proof for compact groups is rather easy: let $x: (\mathcal{H_1}, \pi_1) \to (\mathcal{H_2}, \pi_2)$ be an intertwiner between representations on the compact group $G$. Then if $g \in G$ and $\xi \in \ker (x)$, we have
$$x\pi_1(g)\xi= \pi_2(g)x \xi = 0$$
so that $\pi_1(g)\xi \in \ker(x)$, so the kernel is invariant.

This proof is completely trivial because we have the intertwiner relation $x \pi_1(g) = \pi_2(g)x$ for all $g \in G$. So, one may wonder, if there is a quantum version of this? The answer turns out to be yes. The following facts can all be found in Timmerman's book An invitation to quantum groups and duality (proposition 5.2.7):
If $(H,v)$ is a representation of $(A, \Delta)$, then we can look at the map
$$\pi_v: A^* \to \mathcal{M}(\mathcal{B}_0(\mathcal{H})) = \mathcal{B}(\mathcal{H}): f \mapsto (\text{id}\otimes f)(v).$$  It is shown there that $\mathcal{K}$ is an invariant subspace for $v$ if and only if it is invariant for $\pi_v$.
If $x: \mathcal{H}_1 \to \mathcal{H}_2$ is an intertwiner from $v_1$ to $v_2$, then the relation
$$x \pi_{v_1} (f) = \pi_{v_2} (f)x$$ holds for all $f \in A^*$.
Combining these facts, the proof follows exactly as in the classical case!
