If a language has any single occurrence of a letter, is it not context-free? From what I understand, the rules for CFL from my notes say:

If $L$ is a language and

• for all integers $N$,

• there is a string $w \in L$ of length greater than $N$ such that

• for all ways of writing $w$ as $uvxyz$ with ($v \neq \epsilon$ or $y \neq \epsilon$),

• there is an $i$ such that $u(v^i)x(y^i)z$ is not in $L$,

then $L$ is not context-free.

Does this mean that if we have any equation like: 
$L = \{a^m b a^n \mid m = n\}$, then we can use this for example:
$u=a^m, v=b, x=a, y=a^{n-1}, z=null$

Where we any value of i greater than 1 would make the equation invalid?
This gives us $a^m(b^i)(a)((a^{n-1})^i)$ where $b^2$ means that there are two $b$'s, a string that's not accepted in that language. 

However, in this case, then a Language $L = \{bcde \mid b = c = d = \epsilon\}$ could be proven as not context free. What am I misunderstanding here?
No: context free languages are closed under union, so for any CFL language $L$, we have $L \cup \{ a \}$ is context free, for any letter $a$.

The trick is that, when applying the pumping lemma, we don't get to choose the length $L$.

For a finite language, you just choose $L$ to be larger than any word in the language, and the pumping lemma trivially holds. (Remember, $a \implies b$ is true whenever $a$ is false).

As an intuition, you can see the "length" part of the lemma as being how long of words we need to get to a part of the automaton with loops, or a part of the grammar with recursion. So the small examples will in general not affect the pumping property.