Let $X$ be a locally compact Hausdorff space and $K \in L^{2}(X \times X)$, and $\mu$ a positive borel measure.
Define an operator of $L^{2}(X)$, $T(f(x)) = \int_{X} K(x,y) f(y) \mathrm{d}y$.
Decompose $K(x,y) = \sum_{i=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)}$, where $\psi_{i}(x) = \int_{X} K(x,y) \phi_{i}(y) \mathrm{d}y$, where $\{ \phi_{i}\}$ is an orthonormal basis of $L^{2}(X)$.
Define $K_{N}(x,y) = \sum_{i=1}^{N} \psi_{i}(x) \overline{\phi_{i}(y)}$, and $T_{N}(f) = \int_{X} K_{N}(x,y) f(y) \mathrm{d}y$.
Show that $T_{N}$ is compact.
I am able to show that $T_{N}$ is a finite rank operator, which I believe implies compactness, but I don't know the argument. To see that it is finite rank, we see that $f = \sum_{I=1}^{\infty} c_{i} \phi_{i} \in L^{2}(X)$, then:
$$ T_{N}(f) = \int_{X} \sum_{i=1}^{N} \psi_{i}(x) \overline{\phi_{i}(y)} \sum_{j=1}^{\infty} c_{j} \phi_{j}(y) \mathrm{d}y \\ = \sum_{i=1}^{N} \sum_{j=1}^{\infty} \psi_{i}(x) c_{j} \int_{X} \phi_{j}(y) \overline{\phi_{i}(y)} \mathrm{d}y \\ = \sum_{i=1}^{N} \sum_{j=1}^{\infty} \psi_{i}(x) c_{j} 1_{i=j} \\ = \sum_{i=1}^{N} c_{i} \psi_{i}(x) $$ which implies that $T_{N}$ has its range spanned by $\{ \psi_{i} : i \in [N] \}$. But how do I deduce that $T_{N}$ is a compact operator?
I'll continue what I wrote in my comment. What I'm saying is that you can prove a general statement, not just in your space.
Claim: Let $X$ be any finite dimensional normed space. Then any bounded sequence in $X$ has a convergent subsequence.
Proof: Since in a finite dimensional space all norms are equivalent we can prove it for any other norm. So let's choose a basis $(e_1,...,e_n)$ and define the norm $||x||=\sum_{i=1}^n |c_i|$ where $x=\sum_{i=1}^n c_ie_i$. Ok, so now let's take a sequence $(x_k)_{k=1}^\infty$ and we want to prove it has a convergent subsequence. For every $k$ we can write $x_k=\sum_{i=1}^n c_i^ke^i$. Since by assumption the sequence $||x_k||=\sum_{i=1}^n |c_i^k|$ is bounded we can conclude that in particular the sequence $\{c_1^k\}_{k=1}^\infty$ is a bounded sequence of real or complex numbers. Using Bolzano Weierstrass we get that it has a convergent subsequence $c_1^{k_t}$ which converges to some $c_1$. In a similar way, $(c_2^{k_t})_{t=1}^\infty$ is bounded, hence it has a convergent subsequence $c_2^{k_{t_L}}$ which converges to some $c_2$. Continue this way by induction (every time take a subsequence of the previous sequence of indexes), and at the end we will get a sequence of indexes $\{k_u\}_{u=1}^\infty$ such that $c_i^{k_u}\to c_i$ for all $1\leq i\leq n$. And now it follows that the subsequence $\{x_{k_u}\}$ of the original sequence converges to $\sum_{i=1}^n c_ie_i$.
Ok, now let $X,Y$ be any normed spaces and let $T:X\to Y$ be a bounded linear operator of finite rank, i.e its image has finite dimension. We want to show $T$ is compact. Let $(x_n)$ be any bounded sequence in $X$. Since $T$ is bounded the sequence $(T(x_n))$ must be bounded in $Im(T)$. But $Im(T)$ has finite dimension, so by the claim we proved $T(x_n)$ must have a Cauchy subsequence. So $T$ is compact.