Konig's Theorem states that the cardinality of a maximum matching $M$ in a bipartite graph $G = (U,V,E)$ is equal to the cardinality of a minimum vertex cover $V$.
I have a "proof" of this theorem that seems too simple to work, given other proofs I have seen of this theorem, and I would like to know what could be wrong with this proof. Also, let's supposed I have already proven that for any vertex cover $S$, $|S| \geq |M|$, which is true and not hard to prove.
"Proof":
For simplicity, let us assume $G$ is connected, since if it were not, we can apply the same, following argument to each connected component of $G$.
Going through the vertices in $U$, if $u \in U$ is matched, then we add it to a set $S$ and we add its opposing vertex $v \in V$ to a set $T$. Once we have done that for all vertices in $U$, we proceed to do the same thing for vertices in $V$, as long as the vertex in $V$ being inspected is not already in $T$.
Once we have finished this, we claim that $S$ is a vertex cover of $G$. Suppose it were not. Then there exists an edge $e = (u^\prime,v^\prime) \in E$ such that $u^\prime,v^\prime \notin S$. As a result, neither $u^\prime$ nor $v^\prime$ is matched, or else it would have been added to $S$. Hence, we can match them, adding one more matching to $G$, which contradicts that $M$ is the maximum matching in $G$.
Since each vertex added to $S$ is in bijection with the edges of $M$, $|S| = |M|$. It follows from the result already proven that $|S|$ is a minimum vertex cover.
To illustrate how this isn't sufficient, consider the following matching on $K_{3,2}$:
In this case, you would have $S=\{u_1,u_2\}$ and $T=\{v_1,v_2\}$. S can't be a vertex cover because $u_3$ is not adjacent to any vertex in $S$.