Let $$X=\{f:[0,1]\rightarrow[0,1], f\text{ is Lipschitz continuous}\} $$with the supremum metric .
What can we say about the compactness of$ (X,d).$
I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.
I can't think of anything else. Kindly help !!
On
Posting just for completeness since the answer was already provided, and maybe this comment will be useful for those stumbling upon this question in the future.
A more insightful topology defined on $X$ is the one which present $X$ as a union of compact spaces.
If we call $$ X_n=\{f:[0,1]\to[0,1]\ |\ \mathrm{Lip}(f)\leq n,\ \|f\|_\infty\leq n\} $$ which we know are compact from the Ascoli Arzelà theorem.
The most natural topology on $X=\bigcup_n X_n$ is given by the finest topology which makes all the embeddings $i_n: X_n\to X$ continuous.
The compact subsets of $X$ with this topology are those closed sets which are contained in some $X_n$.
Your metric space $(X,d)$, which we will call $Y$ for clarity, "contains" our space $X$, i.e. we have a continuous mapping $$ X\to Y. $$ This tells us that all the compacts subspaces of $X$ are compacts in $Y$ but a priori $Y$ should have more compacts than just those defined by $X$.
An example of such a compact set given by $$ C=\left\{f_n\in Y\ |\ f_n(t)=\frac{\sin(n^2t)}{n}\right\}\cup\{0\} $$ which converges present a sequence converging uniformly to $0$ but not in any $X_n$.
If $f_n(x)=x^n$ ($x\in[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_{n\in\mathbb N}$ has no convergente subsequence. Therefore, your space is not compact.