I am meant to prove that if $E$ is a finite dimensional subspace of a normed space $X$, then $E$ is a closed subspace.
Now I know that if $E$ is a finite dimensional subspace of $X$, then $E$ is complete. I don't know how completeness of $E$ is different to $E$ being closed.
In both cases we want all limit points to be in $E$, and every limit point comes from a convergent sequence, which means there is a Cauchy sequence going to the limit point, which means that completeness guarantees all limit points are in $E$?
What am I missing?
You have only one way implication, but that's not all.
First of all the concept of closedness requires the set to be embedded in a (larger) space while completeness can be defined from within the set. For example the set $\mathbb Q$ which is not complete is a property not relying on it being part of $\mathbb R$ (that's why it became apparent before introduction of $\mathbb R$). On the other hand $\mathbb Q$ as an topological space in it's own right the entire set will be closed (since it's the complement of the empty set which in turn is open), but as a subset of $\mathbb R$ it's a non-closed set.
This shows also where the reverse implication doesn't hold. In the topological space $\mathbb Q$ we have that it's not complete, yet the set $\mathbb Q$ is closed.
The proof of the implication is as you write. The crusial part where the reverse doesn't hold is that given a sequence $q_n$ converging to $q$ it will apply that it's a Cauchy-sequence.