Theorem: (König's edge-colouring theorem). For any bipartite graph $G=(V, E)$, $$ \chi(G)=\Delta(G) . $$ That is, the edge-colouring number of a bipartite graph is equal to its maximum degree.
Proof. Let $M_1, \ldots, M_{\Delta(G)}$ be a collection of disjoint matchings covering a maximum number of edges. If all edges are covered, we are done. So suppose that edge $e=u v$, say, is not covered. Then (since $\operatorname{deg}(u) \leq \Delta(G)$ ) some $M_i$ misses $u$ and (similarly) some $M_j$ misses $v$. If $i=j$ we can extend $M_i$ to $M_i \cup\{e\}$. If $i \neq j, M_i \cup M_j \cup\{e\}$ makes a bipartite graph of maximum degree at most two. Hence there exist matchings $M$ and $N$ with $M_i \cup M_j \cup\{e\}=M \cup N$. So replacing $M_i$ and $M_j$ by $M$ and $N$, increases the number of edges covered, contradicting our assumption.
Question: There are two things in proof that I did not understand. Firstly, how can we know there exists a collection $M_i$ as above. Secondly, what the statement "replacing $M_i$ and $M_j$ by $M$ and $N$" means?
As you have at least one node with $\Delta(G)$ edges, you can find $\Delta(G)$ disjoint matchings with every single edge coming from this node. Also all those edges cannot be part of the same matching and we then choose the collection of matchings so that no edge is covered twice and still the maximum number of edges is covered.
You have the graph $M_i \cup M_j \cup \{e\}$ but can represent it with a matching $M\cup N$ that covers all edges from $M_i, M_j, \{e\}$. Thus if you replace $M_i, M_j$ in the original collection with $M, N$ you still get a valid collection of disjoint matchings but with one more edge covered. However, this is a contradiction to the original collection being chosen maximal.