Let $T_n$ be the set of nodes on level $n$ of a complete binary tree of height $\omega$: $T_0$ is just the root, so it has $2^0=1$ node, $T_1$ has $2^1$ nodes, and in general $T_n$ has $2^n$ nodes. If the tree isn’t complete, the levels may be smaller, but the point is that they are all finite. Their union is the set of all nodes of the tree, and the union of countably many finite sets is countable.
Now suppose that each node has countably infinitely many children. Label the root of the tree with the empty sequence, $\langle\rangle$. Label the nodes on level $1$, the children of the root, with $1$-tuples of natural numbers: $\langle 0\rangle,\langle 1\rangle,\langle 2\rangle,\dots~$. Label their children with $2$-tuples; for instance, the first three children of $\langle1\rangle$ are $\langle 1,0\rangle,\langle1,1\rangle$, and $\langle1,2\rangle$. On level $4$ you’ll find nodes with labels like $\langle 3,0,5,3\rangle$: this node is the child of $\langle 3,0,5\rangle$, which is the child of $\langle 3,0\rangle$, the child of $\langle 3\rangle$, the child of $\langle\rangle$, the root. In this way you can assign every node in the tree a unique label that is a finite sequence of natural numbers, and every such finite sequence will be attached to a node of the tree. Thus, there are exactly as many nodes as there are labels.
There is only one empty label, and there are obviously $\aleph_0$ (countably infinitely many) labels on level $1$, one for each natural number. The labels on level $n$ are just the elements of the set $\Bbb N^n$, and it’s a basic fact of infinite cardinalities that $\Bbb N^n$ is countably infinite for each $n\in\Bbb Z^+$. Thus, each level of the tree has countably many nodes, and the union of countably many countable sets is still countable, so this tree also has only countably many nodes, not continuum-many.
You went slightly astray with the last sentence: you lost track of the number of internal nodes. If we add the children back in, the number of internal nodes increases by $1$ from $I$ to $I+1$, not from $I+1$ to $I+2$. You also need to show that there actually is an internal node whose children are both leaves. If you’ve already defined the height (or depth, depending on your terminology) of a node, you can pick a leaf $v$ whose height is maximal: its sister must also be a leaf. You can also add a little connective tissue to the induction step, something like this:
Suppose that the result is true for all full binary trees with $I$ internal nodes, and let $T$ be a full binary tree with $I+1$ internal nodes and $N$ nodes altogether. $T$ has at least one internal node, so it must have a leaf; let $v$ be a leaf of maximal height, so that its sister is necessarily also a leaf. Remove $v$ and its sister, and let $T'$ be the resulting tree; the parent of $v$, which was an internal node of $T$, is a leaf of $T'$, and the status of the other nodes of $T'$ is unchanged, so $T'$ has $I$ internal nodes. It is also a full binary tree, so by the induction hypothesis $T'$ has $2I+1$ nodes altogether. Clearly $N=(2I+1)+2=2(I+1)+1$, as desired.
Best Answer
Any full binary tree can be seen as the structure of a single elimination tournament with $n$ teams corresponding to the leaves. Each non-leaf is a game in which the loser goes home and the winner goes up to the next round. There is one final champion, who wins the root game, and so there must be $(n-1)$ non-leaf nodes since every other team loses exactly once. Hence $n + (n-1) = 2n-1$ nodes altogether.