Four Colour Theorem - What was wrong with De Morgan's 'proof' Every resource I can find mentions his 'proof' but then just goes on to other ideas.
Is it because he could not prove his theory?
Surely it is trivial to prove using graph theory notation. (Or don't I know enough about graph theory to see why it isn't?)
 A: In chapter 2, pg. 17 of Four Colors Suffice: How the Map Problem was Solved
by Robin J. Wilson:

In this passage De Morgan hits upon the fact that if a map contains
four regions, each adjoining the other three, then one of them must be
completely enclosed by the others. He believed, incorrectly, that this
idea lay at the heart of the problem, and it soon became an obsession of
his. Since he could not prove it, he proposed to assume its truth as an
axiom, which he defined as “a proposition which cannot be made dependent upon obviously more simple ones.”

Specifically in De Morgan's letter to William Whewell, Master of Trinity College in  Cambridge in December 1853, he wrote:

I soon made out the following—which was at first incredible—then certainly
true—then axiomatic—for I cannot make it depend on anything I see more
clearly.

If four non-interfering compartments have each common boundary line
with the other three—one at least of the four must be inclosed by the other
three—or by fewer . . .

EDIT: @WizardMath, see below:
On pg. 167 of De Morgan on Map Colouring and the Separation Axiom by N. L. Bigg in the Archive for History of Exact Sciences volume 28, pages165–170 (1983), the axiom in modern terms could be summarised as follows:

Axiom. If four regions are mutually adjacent, then one is inclosed.
Theorem. Any plane map requires at most four colours.
Proof. (By induction.) The result is true for any map with not more than four regions. Suppose is it true for maps with at most $n$ regions, and let $M$ be a map with $n+1$ regions.
(i) If $M$ has four mutually adjacent regions, one of them is surrounded by the others. Removing (of shriking) this region results in a map $M'$ with $n$ regions, which can be four-coloured by the induction hypothesis. When the lost region is reinstated it is adjacent to at most three colours, and so there is a colour available for it.
(ii) If $M$ does not have four mutually adjacent regions, four colours are never needed in a 'neighbourhood' and so the induction step proceeds as in (i). Thus the theorem holds for each $n$.

Hope this helps.
