TITLE: Is it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?
QUESTION [18 upvotes]: I wonder about the following statement:

Unit ball is compact in a real Norm Linear Space (NLS) iff the space is finite dimensional.

Is this statement true? How would I go about proving this? I don't want to use that every norm in a finite dimensional real NLS is equivalent because I am using the previous result to prove the later.

REPLY [8 votes]: Is this statement true?
If you are talking about the closed unit ball, yes (as explained in the other answers).
How would I go about proving this?
The proof (of the hardest part) depends on your definition of compactness. If you adopt the sequential definition, the proof follows as explained in the other answers. If you adopt the covering definition, the proof follows as explained below. In the following, $X$ is a normed linear space and $B_X$ stands for the closed unit ball in $X$.

Claim. $B_X$ is compact iff $\dim X<\infty$.

Proof:

($\Rightarrow$) Assume that $B_X$ is compact.

As $\{B(x;\frac{1}{2})\}_{x\in B_X}$ is an open cover of $B_X$, there are $x_1,...,x_n\in B_X$ such that
$$B_X\subset B(x_1;\tfrac{1}{2})\cup \cdots \cup B(x_n;\tfrac{1}{2}).$$ 
As $B(x_i;\tfrac{1}{2})= x_i+\tfrac{1}{2}B(0;1)$ it follows that 
$$B(0;1)\subset B_X\subset Y+\tfrac{1}{2}B(0;1),$$
where $Y=\text{span}\{x_1,...,x_n\}$. Thus
\begin{align*}
B(0;1)\subset Y+\tfrac{1}{2}[Y+\tfrac{1}{2}B(0;1)]&=Y+\tfrac{1}{2^2} B(0;1)\\
&\subset Y+\tfrac{1}{2^2} [Y+\tfrac{1}{2}B(0;1)]=Y+\tfrac{1}{2^3}B(0;1).
\end{align*}
In general,
$$B(0;1)\subset Y+\tfrac{1}{2^n} B(0;1)=Y+B(0,\tfrac{1}{2^n}),\quad \forall\ n\in\mathbb N.$$
So, each $x\in B(0;1)$ has the form $x=y_n+x_n$ with $y_n\in Y$ and $x_n\in B(0;\tfrac{1}{2^n})$. As $x_n\to 0$, we conclude that $y_n\to x$ and thus $x\in \overline{Y}=Y$. This shows that $B(0;1)\subset Y$ which implies that $X\subset Y$ and thus
$$\dim X\leq\dim Y=n<\infty.$$

($\Leftarrow$) Assume that $\dim X=n<\infty$.

There is a topological isomorphism $f:X\to \mathbb K^n$. As $B_X$ is closed and bounded, $f(B_X)$ is closed and bounded. As $\mathbb K^n$ has the Heine-Borel property, $f(B_X)$ is compact. Thus $B_X=f^{-1}(f(B_X))$ is compact. $\square$