Let $X$ be an infinite dimensional vector space over a field $\mathbb{K}$. Suppose that $(X,\|\cdot\|)$ is a complete normed vector space, in the sense that any Cauchy sequence is convergent. Suppose that the closed unit ball of $X$ is compact in the strong topology. 


  
 **Question 1.**
Is $X$ necessarily isomorphic to some finite dimensional Banach space ? 


  
 **Question 2.** If the answer for the question 1 is  **no** , can we always find in $\mathcal{L}(X,X)$ an unbounded linear operator ?


Different from this question [Is there an infinite-dimensional Banach space with a compact unit ball?](https://mathoverflow.net/questions/77750/is-there-an-infinite-dimensional-banach-space-with-a-compact-unit-ball) I would like to assume the axiom of choice. 


**Comment.** 
The example I have in mind is $(\mathbb{R}^n,\|\cdot\|\_{2})$ as $\mathbb{Q}$ vector space. 
Clearly 
$\text{dim}\_{\mathbb{Q}}\ \mathbb{R}^n=\infty$ and the unit ball is compact and this space is complete with respect to the standard Euclidean norm $\|\cdot\|\_2$, but in this example both questions 1 and 2 are trivial. 



Infinite dimensional vector spaces with compact unit ball
I think the following meets your setup. Let $\mathbb K = \mathbb Z\_2$ with the "absolute value" $|0|=0, |1|=1$ (this is non-Archimedean). See <http://en.wikipedia.org/wiki/Absolute_value_%28algebra%29>


Set $V = \mathbb Z\_2^I$ for some index set $I$, with the trivial norm $\|0\|=0$ and $\|x\|=1$ for all other vectors $x$. This satisfies the usual rules, with $\|kx\| = |k|\|x\|$ for $k\in\mathbb K, x\in V$. Clearly $V$ actually have the discrete metric, and so is complete. Now, the closed unit ball is all of $V$, and not compact if $I$ is infinite.


BUT, I could instead define $\|x\|=2$ for $x\not=0$. Still we have a norm. Now the closed unit ball is $\{0\}$; and so is compact. All non-trivial linear maps have norm $1$.


To me, this seems like a very, very silly example, which perhaps shows that the original question needs tweaking with a bit...


If you start with an Archimedean absolute value, then really you have a subfield of $\mathbb C$ or $\mathbb R$ (which must contain $\mathbb Q$). By continuity, I then think you can turn $V$ into a $\mathbb R$ vector space, with a "norm" in the usual sense. Then if $V$ has a compact unit ball, it must be finite dimensional (over $\mathbb R$). So your example of $\mathbb R^2$ over $\mathbb Q$ is in a sense all that can happen.


