$X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min
Here’s an example of how the same argument could be written up nicely.
Since $X$ is compact and $f$ is continuous, $f[X]$ is a compact subset of $\mathbb{R}$ and therefore closed and bounded. Since $f[X]$ is bounded, it has both a supremum and an infimum, and since it is closed, $\sup f[X]\in f[X]$ and $\inf f[X]\in f[X]$. Thus, there are $x_0,x_1\in X$ such that $f(x_0)=\inf f[X]$ and $f(x_1)=\sup f[X]$; that is, $f$ attains its minimum and maximum values at $x_0$ and $x_1$, respectively.
Your argument is fundamentally sound, but you have to assume that your metric space is nonempty. Here is a direct proof that requires no other results (the proof generalizes, like yours, to arbitrary topological spaces):
Let $X$ be a nonempty metric space and $f:X\to\mathbb{R}$ a continuous function without a maximum. Then $X$ has an open cover without a finite subcover.
Proof:
Suppose $f(X)$ is unbounded. Then $\big\{f^{-1}\big((-\infty, n)\big):n\in\mathbb{N}\big\}$ is an open cover without finite subcover.
Suppose $f(X)$ is bounded, with supremum $s$. Since $f$ has no maximum, $s\notin f(X)$ and $\big\{f^{-1}\big((-\infty, s-1/n)\big):n\in\mathbb{N}\big\}$ is an open cover without finite subcover.