Let $a < b$ be real numbers and $f:I:=[a,b] \to \mathbb{R}$ continuous and $f$ having a zero in $I$.
How can one prove that $f$ has a smallest zero, meaning that
$$\min M$$
exists for
$$M := \{x \in [a,b]: f(x) = 0\}$$
Can I say that the set $M := \{x \in [a,b]: f(x) = 0\}$ is the inverse image of $\{0\}$, meaning that $f^{-1}(\{0\}) = M$.
$\{0\}$ is a closed set. Since $f$ is continuous, its inverse image $M$ is closed. In closed sets it holds that $\partial M \in M$, therefore $\min M = \inf M$, so a minimum exists in $M$ (and $M$ was the set of zeros in $f$)
Is that correct?
Expanding slightly on the comment above, because $f^{-1}(\{0\})=M$ is closed, we conclude that $\partial M\subseteq M$. It is important to note that the boundary of a set is itself a set so that $\partial M\in M$ does not really make sense. However, it is true that a closed set contains its boundary points. In particular, because we are working in $\mathbb{R}$, as you say $\inf M= \min M$. As pointed out in the comment above, $\inf M\ge a$ because $a\le x$ for all $x\in M$. So $\inf M \in \mathbb{R}$. Hence $\min M$ exists.