Can you build a solver from a verifier?
Having a verifier for a language in general is known as semidecidability, which is actually weaker than decidability. So in general, the answer is no, we can't build a decider for $L$.
But if the verifier is efficient (i.e., $L \in $ NP), then indeed $L$ is decidable, and your argument is pretty close for why that is.
Let's say we have an efficient verifier $V(x,c)$.
Given an $x \in \Sigma^*$, the &quot;certificate space&quot; is $C=\{c \in \Sigma^* \mid |c| \leq p_L(|x|)\}$, where $p_L$ is a polynomial. That means the certificate space is actually finite. $x \in L$ if and only if there exists a $c \in C$ such that $V(x,c)$ accepts. So indeed we can just check every possible certificate for a given input $x$ to construct a decider for $L$.
In fact with a bit more work you can show that $L \in$ EXP, meaning $L$ is decidable in exponential time.