Compactness of a subspace $X$ is an infinite space, $\mathcal B(X) = \{ f: X \to \mathbb R : \sup|f(x)| < \infty\}$, $d(f,g) := \sup|f(x)-g(x)|$.
I'm trying to debunk that $K := \{f \in \mathcal B(X) : d(f,0) < 1 \}$ is compact.
I already proved that $K$ is restricted and closed.
I am thankful for any help!
 A: I doubt that you can show that $K$ is closed. 
Consider the following example: Let $X=[0,1]$ and consider a sequence $\{f_n\}_{n\ge 1}$ defined over $X$ such that 
$$f_n(x)=\left(1-\frac{1}{n+1}\right)\mathbf 1_{\{1\}}(x), \,\forall\, x\in [0,1].$$
Then it is easy to see that for each $n$
$$d(f_n,0)=\sup_{x\in X}\left|f_n(x)\right|=1-\frac{1}{n+1}<1,$$
which implies that $f_n\in K$ for all $n$. Now consider the function
$$f(x)=\mathbf 1_{\{1\}}(x), \,\,\forall\, x\in [0,1].$$ 
Notice that 
$$d(f_n, f)=\sup_{x\in X}|f_n(x)-f(x)|=\frac{1}{n+1}\to 0\,\,\text{as }n\to\infty,$$
meaning that $f_n$ converges to $f$ under the sup-norm. However, it is easy to see that $d(f,0)=1$ and hence $f\notin K$.
Therefore, $K$ is not closed, and hence it is not compact.
A: I presume that by "restricted" you mean what is usually called "bounded".
How about this set of functions: For each $y\in X$ we have a function
$$
f(x) = \begin{cases} 1/2 & \text{if } x = y, \\
0 & \text{if } x\ne y. \end{cases}
$$
See if you can prove that this is a closed subset of $K$ and has an open cover that has no finite subcover. An open ball of small radius about each of these functions ought to do it.
