I am reading the following lemma from Munkres' Elements of Algebraic Topology:
Lemma 2.1 A collection $K$ of simplices is a simplicial complex if and only if the follow hold:
Every face of a simplex of $K$ is in $K$.
Every pair of distinct simplices of $K$ have disjoint interiors.
Now I am trying to follow the proof that if 1 and 2 above hold then $K$ is a simplicial complex. Although munkres does not define what he means by "distinct simplices", I am told by Mixedmath that this means two simplices that don't have a common vertex. The proof according to Munkres goes as follows:
Let $\sigma$ and $\tau$ be two distinct simplices of $K$ such that they have disjoint interiors. Let $\sigma'$ be the face of $\sigma$ that is spanned by those vertices $b_0,\ldots,b_m$ of $\sigma$ that lie in $\tau$. The claim now is that $\sigma \cap \tau$ is equal to $\sigma'$. Now one direction I understand the other which I don't is when he shows that
$\sigma \cap \tau \subseteq \sigma'.$
The line I don't understand is this:
Let $x \in \sigma \cap \tau$. Then $x \in \textrm{Int}\, s \cap \textrm{Int} \hspace{2mm} t$ for some faces $s$ of $\sigma$ and $t$ of $\tau$.
How does this follow from the assumption of (2) above?