Embedding of Tree Q. 
Proof for every Tree can be embedded into the plane.
Conditions. 
We cannot use Euler Formula for Planar Graphs. We can use definition of tree, $V-E=1$, no-cycles, every edge is critical, there is a vertex of frequency at most one.
Attempt.
We prove this by induction. True for $V$=1. Trivial. Consider true for $V<n$. Consider a tree $T$ with $n$ edges. Now, we know it has a vertex with at most one edge incident in it. Consider $T$\ $\{v,l\}$ the last vertex and edge. This can be embedded by inductive hypothesis. For the $v$ and $l$, that is left, it can be easily embedded. 
The last part seems sloppy. I would appreciate other proofs or improvement on this proof.
 A: You may prefer to chose an embedding where all edges are mapped to straight line segments.
As you did, let $v$ be a vertex of degree $1$, let $w$ be its only neighbour.
As you are not allowed to use very much, I assume you already know/can use that $v\ne w$ and removing vertex $v$ and edge $vw$ from $T$ does indeed produce a tree $T'$.
By induction hypothesis, there is an embedding $f\colon T\to\mathbb R^2$ of $T'$ where all edges are mapped to straight line segments.
Especially, the image of any edge $xy$ where $w\ne\{x,y\}$ is compact, hence has positive distance to $f(w)$.
Since there are only finitely many such edges, for some $r>0$ the open ball $B_r(f(w))$ interscts only the finitely many straight line segments $f(wx)$ for edges incident with $w$.
Select $f(v)$ in $B_r(f(w))\setminus\bigcup_{x\text{ neigbour of } w}f(wx)$ and map $vw$ to the straight line segment from $f(v)$ to $f(w)$.
A: You're right that there's something very subtle missing.  Suppose we have a tree with uncountably many straight edges, from $(0,0)$ to every point on the circle $x^2+y^2=1$.  Now you can't add another edge to $(0,0)$.  However if your tree is finite your proof is fine.
