You could stand to give a bit more detail. Let $\tau_Y$ be the subspace topology on $Y$ and $\tau$ the order topology on $Y$, and for $a,b\in Y$ write $(a,b)_Y$, $(a,b]_Y$, etc. for intervals in the relative order on $Y$. In showing that $\tau\subseteq\tau_Y$, you can specify precisely how the intervals in $Y$ are obtained from intervals in $X$.
- If $a,b\in Y$ with $a<b$, then $(a,b)_Y=(a,b)\cap Y\in\tau_Y$.
- If $a_0=\min Y$, and $b\in Y\setminus\{a_0\}$, then either $a_0=\min X$, in which case $[a_0,b)_Y=[a_0,b)\cap Y\in\tau_Y$, or there is an $a\in(\leftarrow,a_0)$, in which case $[a_0,b)_Y=(a,b)\cap Y\in\tau_Y$.
- The case in which $b_0=\max Y$, $a\in Y\setminus\{b_0\}$, and we consider $(a,b_0]_Y$ is similar.
For the proof that $\tau_Y\subseteq\tau$, note that it is not true that every open set in $Y$ is of the form $(c,d)\cap Y$ for some $c,d\in X$ with $x<d$. It isn’t even quite true that these sets form a base for $\tau_Y$: that’s the case if and only if $X$ has neither a minimum nor a maximum element. It’s fine to work with a base for $\tau_Y$, of course, but you do have to split it into three cases, just as in the other part of the proof. Moreover, since $c$ and $d$ need not be in $Y$, even showing that $(c,d)\cap Y\in\tau$ takes a bit of work that you’ve not supplied.
Let $c,d\in X$ with $c<d$, and let $B=(c,d)\cap Y\in\tau_Y$. If $c,d\in Y$, then the convexity of $Y$ ensures that $B=(c,d)_Y\in\tau$, but there’s no reason to assume that $c$ and $d$ are in $Y$. To show that $B\in\tau$, we’ll show that for each $y\in B$ there is a $U_y\in\tau$ such that $y\in U_y\subseteq B$; $B$ will then be a union of members of $\tau$ and hence itself in $\tau$. Let $y\in B$; clearly $c<y<d$.
If $y$ is neither the minimum nor the maximum element of $B$, there are $u,v\in B$ such that $c<u<y<v<d$. $Y$ is convex, and $u,v\in Y$, so $(u,v)\subseteq Y$, and therefore $(u,v)_Y=(u,v)$. Thus, $y\in(u,v)_Y\subseteq B$, where certainly $(u,v)_Y\in\tau$.
If $y=\min B$, and there is some $v\in B$ such that $y<v$, there are two possibilities. One is that $y=\min Y$; in that case $[y,v)\subseteq Y$, so $y\in[y,v)_Y\subseteq B$, and $[y,v)_Y\in\tau$. The other is that $y\ne\min Y$, in which case there is some $u\in Y$ such that $u<y$. Clearly $u\le c$, since $y=\min B$, and the convexity of $Y$ then implies that $c\in Y$. But then $y\in(c,v)_Y\subseteq B$, and $(c,v)_Y\in\tau$. (It’s not hard to show that if $y\ne\min Y$, then in fact not only is $c\in Y$, but $c$ is the immediate predecessor of $y$ in $X$: the open interval $(c,y)=(c,y)_Y=\varnothing$.)
If $y=\max B$, and there is some $u\in B$ such that $u<y$, you can show similarly that either $y=\max Y$, and $y\in(u,y]_Y\subseteq B$ with $(u,v]_Y\in\tau$, or $d\in Y$, and $y\in(u,d)_Y\subseteq B$ with $(u,d)_Y\in\tau$.
The only remaining possibility is that $y=\min B=\max B$, so that $B=\{y\}$. In this case you can use the same ideas to show that $c,d\in Y$, so that $B=(c,d)_Y\in\tau$.
It really is easier to work with the subbase of open rays: there are fewer cases that need to be handled separately.
