Open Sets in Subspaces of a Metric Space I seek some clarification in a step of the following proof of this theorem:

Theorem. Let $M$ be a metric space and $(X_{\lambda})_{\lambda\in L}$ a family of connected subsets of $M$ which contain a common point $a\in M$. Then $X = \bigcup\limits_{\lambda\in L}X_{\lambda}$ are connected.

Beginning of the proof: Let $A\cup B$ a union of disjoint open subsets of $M$ such that $A\cup B = X$. The point $a$ is in one of either $A$ or $B$, say it's in $A$. For all $\lambda\in L$, $A\cap X_{\lambda}$ and $B\cap X_{\lambda}$ are both open in $X_{\lambda}$. Then $X_{\lambda}=(A\cap X_{\lambda})\cup(B\cap X_{\lambda})$. [...] 
I didn't get why the bold part is true, could someone please explain it further?

For each $\lambda$, $X_{\lambda}$ has the subspace topology, so the open sets in $X_{\lambda}$ are $U\cap X_{\lambda}$, where $U$ is open in $M$. This is the definition of the subspace topology.
