Compact operators and dimension If every compact operator is of finite dimensional range,does it follow that the dimension of the space is finite?
 A: If your underlying space $X$ is a Banach space, the answer is "Yes".
To prove this, assume that $X$ is infinite-dimensional, and let us show that there exists a compact operator on $X$ whose range is infinite-dimensional.
It is a standard fact that since $\dim(X)=\infty$, one can find a biorthogonal sequence $(e_n,e_n^*)_{n\in\mathbb N}$ in $X\times X^*$, where $X^*$ is the dual space of $X$. This means that we have $e_n^*(e_m)=0$ if $n\neq m$ and $e_n^*(e_n)=1$.
Let $(\alpha_n)_{n\in\mathbb N}$ be a sequence of positive numbers such that $C:=\sum_1^\infty \alpha_n \Vert e_n\Vert\,\Vert e_n^*\Vert<\infty$. Then the formula 
$$T(x):=\sum_{n=1}^\infty \alpha_n e_n^*(x) e_n$$
makes sense for every $x\in X$ because $X$ is a Banach space and $$\sum_{n=1}^\infty \Vert\alpha_n e_n^*(x)e_n\Vert\leq\sum_{n=1}^\infty \alpha_n \Vert e_n^*\Vert\,\Vert e_n\Vert\, \Vert x\Vert=C\,\Vert x\Vert<\infty\, .$$
Moreover, we have $\Vert T(x)\Vert\leq C\, \Vert x\Vert$ for all $x$, so $T$ is a bounded operator on $X$.
For any $k\in\mathbb N$, we have $T(e_k)=\varepsilon_k e_k$ by the "biorthogonality" condition. So the range of $T$ contains all vectors $e_k$, and hence it is infinite-dimensional because these vectors are linearly independent (again by the biorthogonality condition).
Finally, if we set $T_N(x):=\sum_{n=1}^N e_n^*(x)\, e_n$, then 
$$\Vert T(x)-T_N(x)\Vert\leq \left(\sum_{n>N} \alpha_n \Vert e^*_n\Vert\, \Vert e_n\Vert\right) \Vert x\Vert:=\varepsilon_N \,\Vert x\Vert $$
for all $x\in X$. So $\Vert T_N-T\Vert\leq\varepsilon_N$ and hence $T_N\to T$ in the operator norm. Since the $T_N$'s have finite dimensional ranges, it follows that $T$ is compact.
