Let $A_N: L^2([0,2\pi])\to L^2([0,2\pi])$ be given by
$$ f(x) \mapsto (A_Nf)(x) = \sum_{n=-N}^N \hat f(n) e^{inx} $$ where $$ \hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} f(x) e^{-inx} dx $$
For, say $N=0$, we have
$$ (A_0f)(x) = \hat f(0), $$ which is just a constant, i.e. the image has dimension one. So $A_0$ is a finite rank operator and so it is compact. And similarly for any value of $N$ we take, $A_N$ will be a finite rank operator and hence compact.
Now define the operator $A: L^2([0,2\pi])\to L^2([0,2\pi])$ by
$$ f(x) \mapsto (Af)(x) = \sum_{n=-\infty}^\infty \hat f(n) e^{inx}. $$ This operator is clearly the limit of the sequence of finite rank operators $\{A_N\}$ so it is compact. However, it is also just the identity operator as it simply gives the Fourier decomposition of $f$. I.e. $Af = If = f$. But it holds that the identity operator, and hence $A$, is not compact in an infinite dimensional space.
So where have I gone wrong? Is $A$ compact or not?
The theorem that the limit of a sequence of finite-rank operators (or compact operators) is compact refers to the norm topology on the space of continuous linear maps. If $(T_n)$ is a sequence of compact operators in $L(X,Y)$ (where $Y$ is a Banach space, $X$ a normed space) and $T_n \to T$ in the norm topology of $L(X,Y)$, then $T$ is compact.
The sequence $(A_N)$ is not norm-convergent. We have $A_N(f) \to f$ in the norm topology of $L^2([0,2\pi])$, which means that $A_N \to \operatorname{id}$ in the strong operator topology, but that is a much weaker topology than the norm topology on $\mathscr{B}(L^2([0,2\pi]))$.