Lemma 2.2 If $L$ is a subcomplex of $K$, then the polytope $\lvert L \rvert$ is a closed subspace of the polytope of $K$, denoted $\lvert K \rvert$. In particular, if $\sigma \in K$, then $\sigma$ is a closed subspace of $\lvert K \rvert$.
Proof: Suppose $A$ is closed in $\lvert L \rvert$. If $\sigma$ is a simplex of $K$, then $\sigma \cap \lvert L \rvert$ is the union of those faces $s_i$ of $\sigma$ that belong to $L$. Since $A$ is closed in $\lvert L \rvert$, the set $A \cap s_i$ is closed in $s_i$, and hence closed in $\sigma$. Since $A \cap \sigma$ is the finite union of the sets $A \cap s_i$, it is closed in $\sigma$. We conclude that $A$ is closed in $\lvert K \rvert$. Conversely, if $B$ is closed in $\lvert K \rvert$, then $B \cap \sigma$ for each $\sigma \in K$, and in particular for each $\sigma \in L$. Hence $B \cap \lvert L \rvert$ is closed in $\lvert L \rvert$.
I am using Elements of Algebraic Topology by Munkres and it does not give the definition of a closed subspace. It also does not appear to be using the definition here: What is a "closed subspace" of a topological space?.
To sum up what we have said in the comments. Munkres is indeed using the same definition of a closed subspace, in this case, with respect to the weak topology: a subspace $C$ of $|K|$ is closed if and only if $C\cap \sigma$ is closed in $\sigma$ for every simplex $\sigma$ in $|K|$. So he shows that this is true for $|L|$ by showing it for every closed subspace $A$ of $|L|$ (in particlar, for $|L|$ itself).
In addition, in the last line he shows that the subspace topology of $|L|$ in $|K|$ is the same as the weak topology of $|L|$, so $|L|$ is naturally a subspace of $|K|$.
Finally, to also add the answer to your follow up question, $s_i$ stands for a (not necessarily top-dimensional) face of $\sigma$. Since a simplex has finitely many faces, then we can guarantee that there are finitely many $A\cap s_i$.
The reason why an $n$-simplex has finitely many faces is a consequence of the definition of a face. A face of an $n$-simplex consists of those points $(t_0,\dots, t_n)$ such that at least one $t_i=0$. There are only finitely many possibilities for that.