A number of sources state that the following:$^\ast$ that if $C$ is an $R$-coalgebra ($R$ a commutative ring), with $M$ a right $C$-comodule, and $\iota:K\to M$ a $C$-pure submodule, such that $\rho_M(K)\subseteq K\otimes C$, then $K$ inherits a right $C$-comodule structure from $M$. Indeed if $\rho_M:M\to M\otimes_RC$ is the coaction, then you obtain a (unique) map $\rho_K:K\to K\otimes_RC$ such that $\iota\otimes\text{Id}_C\circ \rho_K=\rho_M\circ \iota$. My question is how you show that this is coassociative. One can obtain \begin{equation*} (\iota\otimes\text{Id}_C\otimes\text{Id}_C)\circ (\rho_K\otimes\text{Id}_C)\circ \rho_K=(\iota\otimes\text{Id}_C\otimes\text{Id}_C)\circ (\text{Id}_K\otimes\Delta)\circ \rho_K \end{equation*} and then to finish the argument, you need that $\iota\otimes\text{Id}_C\otimes\text{Id}_C$ is monic - so you need that $\iota$ is $(C\otimes C)$-pure. Does this follow from it being $C$-pure? Or is there some other way to establish coassociativity of $\rho_K$?
${}^\ast$ For instance "Corings and Comodules" by Brzezinski and Wisbaurer, p. 24; https://d-nb.info/980423562/34, p. 45; and "Mathematics & Mathematics Education: Proceedings of the Third International Palestinian Conference", p. 24
Edit: Above I suggested a possible proof, which would go through if $\iota:X\to Y$ being $C$-pure implied it was $C\otimes C$-pure. We now give an example to show that this is false in general. We note two general facts: firstly, that if $C=\bigoplus_{i\in I}C_i$ is a direct sum, then $f$ is $C$-pure iff it is $C_i$ pure for each $i$; and secondly, that an inclusion $\iota:X\to Y$ is $\frac{R}{I}$ pure iff the map $$\frac{X}{IX}\to \frac{Y}{IY},\,\,\,x+IX\mapsto \iota(x)+IY$$ is injective. Now we take $R=\mathbb{Z}[x,y]$, $C=\frac{R}{I}\oplus \frac{R}{J}$, $X=(x+y)\subseteq R$, $Y=R$. Then if $f\in (x+y)\cap IY$, then we have $(x+y)|f$ and $x|f$ and so $x(x+y)|f$ since $R$ a UFD, and so $f\in IX$. Thus the map $\frac{X}{IX}\to \frac{Y}{IY},\,x+IX\mapsto \iota(x)+IY$ is injective, so that $\iota$ is $\frac{R}{I}$-pure. Similarly $\iota$ is $\frac{R}{J}$ pure, and so $C$-pure.
Now since $\frac{R}{I}\otimes\frac{R}{J}\cong \frac{R}{I+J}$ in general, we have $C\otimes C\cong \frac{R}{I}\oplus \frac{R}{J}\oplus \frac{R}{I+J}\oplus\frac{R}{I+J}$. Thus $\iota$ is $C\otimes C$-pure iff it is $\frac{R}{I}$-pure, $\frac{R}{J}$-pure, and $\frac{R}{I+J}$-pure. We have seen that the first two hold. However we have $x+y\in I+J$, so $X\subseteq (I+J)Y$, so the map $$g:\frac{X}{(I+J)X}\to\frac{Y}{(I+J)Y},\,\,\,x+(I+J)X\mapsto \iota(x)+(I+J)Y$$ is the zero map. Then every element of $(I+J)X=(x(x+y),y(x+y))$ has every nonzero monomial of degree $\geq 2$, so that $(x+y)\notin (I+J)X$, and thus $\frac{X}{(I+J)X}\neq 0$. Thus $g$ is not injective, so that $\iota$ is not $C\otimes C$-pure, as needed.
In the errata (http://reh.math.uni-duesseldorf.de/~wisbauer/corinerr.pdf) to one book where this is stated ("Corings and Comodules" by Brzezinski and Wisbaurer), the authors correct the claim to the claim that if $\iota$ is $C\otimes_RC$ pure then the result follows. So it appears that the claim in the question is just a false claim that has been repeated in the literature.