You misunderstood something. Since $\tau$ only includes subsets of $X$. It does not include $\mathcal P(X)$ itself as an element.
Instead it includes every element of $\mathcal P(X)$ (read: subset of $X$) which can be written as the union of these singletons. Which subsets are these?
Yes, this is correct. Every subset of $X$ is open in this case.
As for the final question, yes, it is possible to have bases of different cardinalities, for example by taking $\{\{a\},\{b\},\{c\},\{a,b\}\}$ you obtain another basis for the same topology, and of course the topology itself is always a basis for itself.
It looks to me as if you have a fundamental misunderstanding of some kind. In your answer to (b) you write:
Also, the largest coordinate 1 x 1 does not have a neighborhood in Y, since the x-coordinate of 1 x 1 is not in Y.
This doesn’t make sense: $Y$ is a set of ordered pairs of real numbers, so individual coordinates of those ordered pairs can’t possibly be elements of $Y$. In any case the point $1\times 1$, or $\langle 1,1\rangle$ as I prefer to write it, isn’t in $Y=(0,1)\times[0,1]$ anyway, since its first coordinate isn’t in $(0,1)$, so whether $\langle 1,1\rangle$ has an open nbhd in $Y$ is irrelevant.
As it happens, $(0,1)\times[0,1]$ is open. For each $x\in\left(0,\frac12\right)$ try to sketch the open interval
$$\left(\left\langle x,\frac12\right\rangle,\left\langle 1-x,\frac12\right\rangle\right)\;,$$
or in your notation
$$\left(x\times\frac12,(1-x)\times\frac12\right)\;;$$
can you show that every point of $Y$ is in one of these open intervals, and that each interval is a subset of $Y$?
In (a) the set $Y=(0,1)\times(0,1)$ is indeed open, but it’s not equal to $(0\times 0,1\times 1)$: $(0\times 0,1\times 1)$ is all of $X$ except the two endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$. HINT: Note that for each $x\in(0,1)$ the interval
$$\big(\langle x,0\rangle,\langle x,1\rangle\big)=(x\times 0,x\times 1)$$
is a subset of $Y$.
In (c) you’ve made the same error as in (b): it makes no sense to ask whether a coordinate of a point is in $Y$, and anyway the point $\langle 0,0\rangle$ isn’t in $Y$ in the first place. Do (a) first; once you’ve done it, this one should be pretty straightforward.
Best Answer
The general way to contruct a base for an ordered space is the following:
If $(X,<)$ is an ordered space, a subbase for its order topology is given by all sets of the form $L(a) = \{x \in X: x < a\}, a \in X$ (the lower sets) together with all sets $U(a) = \{x: x > a\}$, where $a \in X$ (the upper sets).
The base derived from it (i.e. all finite intersections of subbase elements) depends on whether $X$ has a minimum $m$ or maximum $M$, as $m$ cannot be in any upper set, nor $M$ in any lower set. All others can. $L(a) \cap L(a') = L(\min(a,a'))$ and $U(a) \cap U(a') = U(\max(a,a')$ so both types of sets among themselves are closed under finite intersections. So we only need to consider the intersections of one $L(a)$ and one $U(a')$ which are exactly the open intervals $(a',a)$ in this case, and we only have possibly non-empty intersections when $a' < a$. If we have $m$ we also need to include all $L(a)$ in the base (where $a>m$, and if we have $M$, all $U(a)$(where $a < M$ as well.
So for $I \times I$ in the lexicographic order $<_l$ we do have $m= 0 \times 0$, and $M = 1 \times 1$, so the standard base is indeed by the above general procedure:
Which is the same as the basis you describe.