Could someone point out to me where the error(s) are in my reasoning below?
First-order logic (with the sequent calculus as the proof system) is both sound and complete. This is the same as saying the set $S$ of semantic consequences of a theory (the statements that are true in every model for which the statements of the theory are also true) is the same as the set $T$ of its syntactic consequences (using the sequent calculus). This is because soundness means that $T$ is a subset of $S$ and completeness means that $S$ is a subset of $T$, i.e. that $S = T$.
This means that every theory is consistent: it is clear that $S$ cannot contain both a statement and its negation (since a model can only satisfy one of the two), but by the above the same must be true for $T$, so that the theory is consistent.
Best Answer
The mistake in your argument occurs (as usual) when you write "it is clear...".
Here, to conclude that $S$ cannot contain both $\varphi$ and $\lnot \varphi$, you implicitly assume that your theory has a model.
Recall that for a theory $\Sigma$, the set $S$ of semantic consequences of $\Sigma$ is $$S = \{\varphi\mid \forall M\,(M\models \Sigma\to M\models \varphi)\}.$$ If $\Sigma$ has no models, then for every sentence $\varphi$, $\varphi\in S$ (vacuously). In particular, both $\varphi$ and $\lnot \varphi$ are in $S$, and hence also in $T$. So $\Sigma$ is inconsistent.
What the argument above shows is that if $\Sigma$ has no models, then $\Sigma$ is (syntactically) inconsistent. The contrapositive is that every consistent theory has a model - which, incidentally, is the main way the completeness theorem is used in model theory.
The converse, that an inconsistent theory has no models, is clear. So we have that the consistent theories are exactly those with no models (and the consistent theories are exactly those with models).