showing that the untwisted Whitehead double of a slice knot is slice

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Let $K$ be a slice knot (i.e. there is a smooth embedding of a disc into $D^4$ such that its boundary is precisely $K$). To obtain the positive (or negative) untwisted Whitehead double of $K$, push a copy $K'$ of $K$ slightly into a Seifert surface of $K$ and add a positive (or negative) clasp as shown here for example for the trefoil: https://www.researchgate.net/figure/A-Whitehead-double-of-a-trefoil-knot-is-a-satellite-knot-whose-companion-knot-is-a_fig10_339015556

Now if the starting knot is slice (this is not the case for the trefoil), then the knot resulting from this construction is slice. Here is how I think one can show this:

Since $K$ and $K'$ are both slice, they bound smooth discs in $D^4$. If I could isotope these surfaces (with their boundaries staying in S^3) such that they do not intersect, I could just attach a twisted band in $S^3$ from one knot boundary to the other, such that the result would be the creation of the clasp. The resulting surface would be a disc since it was obtained my attaching a band to two disjoint discs. But I am stuck on the first part of showing that I can make both slice discs disjoint. I know that one has to use the fact that $K'$ is obtained from $K$ by pushing it inside a Seifert surface, else their linking number would be non-zero which obstructs the discs from being disjoint by Lemma 8.13 in "An Introduction to Knot Theory" by Lickorish (The lemma states precisely that for two disjoint slice discs of two knots, their linking number is $0$). So here is my question:

How can I get the link $K\cup K'$ to bound two disjoint smoothly embedded discs in $D^4$?

Somehow I think one should just be able to shrink the slice disc of $K$ into a slice disc of $K'$ so that they do not intersect. What I worry about here is a kind of phenomenon where two surfaces in a 4-manifold will always intersect, like for $CP^1 \subset \mathbb{C}P^2$. So maybe I can argue by saying something about the trivial intersection form of $D^4$ or $S^4$? Most likely there is a way simpler answer, but I am also curious about when in general, one can push a surface away from itself to make it disjoint.

Thank you for your time.