Proof of Theorem 16.4 in Munkres Topology 2nd Edition For some reason, I cannot make sense of a statement in this proof. The statement is: 
"Since the sets $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ form a subbasis for the subspace topology on $Y$, and since each is open in the order topology, the order topology contains the subspace topology."
I understand these sets form a subbasis for the subspace topology on $Y$, and I understand that each is open in the order topology. However, I cannot understand how these conditions lead to the implication, that the order topology contains the subspace topology. 
If I denote the order topology $\tau_o$ and the subspace topology $\tau_s$, and I denote the subbasis for the subspace topology $\mathcal{S}_s$, then, after some thought, I get:
$U \in \mathcal{S}_s \implies U \in \tau_s$, since $\mathcal{S}_s$ is a subbasis for $\tau_s$, so $\mathcal{S}_s \subset \tau_s$
and 
$U \in \mathcal{S}_s \implies U \in \tau_o$, since elements of the $\mathcal{S}_s$ are open in the order topology, so $\mathcal{S}_s \subset \tau_o$. 
But from this, I obviously cannot conclude that $\tau_s \subset \tau_o$. What am I missing? 

You have confused yourself with notation. You don't care about whether every subbasis element is in both topologies. To say that $\tau_s\subset\tau_o$ is to say that every open set in the subspace topology is open in the order topology. But every subbasis element for the subspace topology is open in the order topology and hence all finite intersections of subbasis elements are as well. Thus, every open set in the subspace topology is open in the order topology, as we wished.
