If $X$ is a compact space and $Y$ a metric topological space with metric $d$, then $d'(\alpha,\beta) = sup_{x \in X} d(\alpha (x),\beta (x))$ is a metric on $TOP(X,Y)$
$TOP(X,Y)$ denotes the set of continuous functions from $X$ to $Y$ and the definition of compact I have been given is that any open cover has a finite subcover.
My question is why is the supremum above guaranteed to exist? My guess is its due to the compactness of $X$ but I don't understand how exactly.
A sketch for a direct proof: for $x \in X$ we can can apply continuity of both $\alpha$ and $\beta$ to find some open neighbourhood $U_x$ of $x$ so that
$$\forall x' \in U_x: d(\alpha(x'), \beta(x')) < d(\alpha(x), \beta(x)) + 1$$
Now we have a finite subcover $U_{x_1}, \ldots, U_{x_n}$ of the cover so-defined and this allows us to bound $d'(\alpha, \beta)$ on $X$..