Is it "obvious" that, for any fields $K\subseteq L \subseteq M$ and any subset $S \subseteq M,$ one has $$[L(S):L]\leq [K(S):K]?$$
Thinking so far...
This feels "obvious", for example if $S=\{\alpha\},$ then we have three easy cases:
1) $\alpha$ is transcendental $/K$ and algebraic over $/L.$
In this case, $[L(S):L]$ is finite and $[K(S):K]$ is equal to $\aleph_0,$ so the result holds.
2) $\alpha$ is transcendantal $/L.$
In this case, $[L(S):L]$ and $[K(S):K]$ both equal to $\aleph_0,$ so the result holds.
3) $\alpha$ is algebraic $/K.$
In this case, $m_{\alpha/L}$ divides $m_{\alpha/K}$ in $L[X]$ and so $$[L(S):L]=\deg(m_{\alpha/L})\leq\deg(m_{\alpha/K})=[K(S):K].$$
So we can probably prove the result when $S$ is finite by induction...
Question. But what about when $S$ is infinite?
If $[K(S):K]$ is finite then $K(S)=K(S')$ with $S'$ finite.
Do you mean comparing the cardinality when $[K(S):K]$ is infinite ?
If $L/K$ is algebraic then $L(S)/K(S)$ is algebraic so $L(S)=K(S)(L)=K(S)[L]$ is a quotient of $K(S)\otimes_K L$ and hence $$K(S)=\bigoplus_{b\in B} b K \implies L(S)=\sum_{b\in B} bL=\bigoplus_{b\in B'} bL$$ with $B'\subset B$ so $$[L(S):L]=|B'|\le |B|=[K(S):K]$$
Similarly if $K(S)/K$ is algebraic then $L(S)=L[K(S)]$ is a quotient of $L\otimes_K K(S)$ so that $K(S)=\bigoplus_{b\in B} b K \implies L(S)= \sum_{b\in B} bL$ and hence $[L(S):L]\le[K(S):K]$
If none of $L/K$ and $K(S)/K$ are algebraic then it doesn't have to hold anymore: $[\Bbb{Q}(t):\Bbb{Q}]$ is countably infinite whereas $$[\Bbb{C}(t):\Bbb{C}]= [\bigoplus_{n\ge 0} t^n \Bbb{C}\bigoplus_{a\in \Bbb{C},m\ge 1} \frac1{(t-a)^m}\Bbb{C}:\Bbb{C}]$$ is not.