There is a big difference between finite-dimensional and infinite-dimensional spaces. We know for example that the unit ball in a normed vector space is compact if and only if the vector space is finite-dimensional.
But we are most familiar with finite-dimensional spaces. We are happy if our operator at hand has at least finite-dimensional range (we call them finite-dimensional operators), because we have a good feeling for these. Now, compact operators are in a sense closest to finite-dimensional operators as you can get. For example, in a Hilbert space every compact operator is the limit of a sequence of finite-dimensional operators.
You also get some feeling for this when you consider the spectrum of an operator. Let $A$ be a bounded operator from a Banach space $X$ into itself. Its spectrum is defined as the set of all complex numbers $z$ for which $A - zI$ is not bijective. In finite dimensions this is just the set of eigenvalues, right? The spectrum of a general bounded operator is always compact. However, it can be rather weird. A simple example is the multiplication operator $T : L^2(0,1)\to L^2(0,1)$ mapping a function $f$ to $xf(x)$. You can easily compute its spectrum. It is the whole interval $[0,1]$, but it has no eigenvalues at all. However, when $A$ is a compact operator all points in its spectrum (except $z=0$) are eigenvalues and you have have Jordan blocks as in the finite-dimensional situation. The only difference is that there might be infinitely many eigenvalues (which then accumulate to $z=0$) and that $z=0$ might be not an eigenvalue, although it is always a spectral point. There are also weird compact operators like the Volterra operator whose only spectral point is $z=0$, which is not an eigenvalue.