show that all trees are planar without using Euler's formula or the fact that all trees are planar

This proof does technically use "induction," but it should be phrased in a way that a layperson will still be able to understand the mechanics behind the argument. If your goal is to explain why something is true intuitively instead of mathematically, words like "induction" can be confusing, but induction is a fairly intuitive concept.

In this proof I'll be using the definition of a tree as a "connected graph with no cycles."

Draw the tree one edge at a time. Each iteration we'll pick a new edge to draw, and figure out a way to draw it. We'll pick the edge in a particular way -- we will pick an edge that shares an endpoint with something we've already drawn. The reason we can do this is that, if at some point we run out of edges that share an endpoint with something we've already drawn, then we must be out of edges completely -- for any other edge that exists, there must be some path in the tree from it to some vertex we've already drawn (connectedness), and in that path there has to be some first edge that we haven't already drawn.

I claim that, in fact, this edge has only one endpoint that we've already drawn. This is true because, if it were to share both endpoints, then this would create a second path between two already-drawn vertices. Because every time we draw an edge we make sure it has a previously-marked endpoint, our tree remains connected whenever we draw anything. So, for any two vertices we've drawn, there already exists a path between them -- if the tree had an edge between them, this would give it a cycle.

Now, all we need to show is that we can actually draw it. Look at the endpoint, and consider some angle coming out of it where there isn't already an edge. Draw our new edge at that angle, and make it really really small.