Suppose $\tau$ is such a topology. For clarity, I'll use "open" to mean open in the usual sense, and "$\tau$-open" to mean open in the new topology, $\tau$; similarly with "continuous" and "$\tau$-continuous."
We'll show that $\tau$ is in fact a refinement of the usual topology on $\mathbb{R}$. The only fact about the usual topology used is $$(*)\quad\mbox{No map $h:\mathbb{R}\rightarrow\mathbb{R}$ whose range has exactly two elements is continuous.}$$
We begin by showing that $\tau$ is $T_1$. Take $a\neq b$, and take two non-empty subsets $A,B$ with $A\cap B=\varnothing$ and $A\cup B=\mathbb R$ and with $B$ $\tau$-open. Such $A, B$ exist since $\tau$ is not indiscrete: if $\tau$ were, it would have too many continuous functions.
Now by $(*)$, the function that sends $A$ to $a$ and $B$ to $b$ is not continuous in the usual topology, so it is not $\tau$-continuous either.
Since $f$ isn't $\tau$-continuous, we must have that $A$ is not $\tau$-open. Now, the only $f$-preimages are $\emptyset, B, \mathbb{R}, A$; the only one of these which is not $\tau$-open is $A$, so for $f$ to not be $\tau$-continuous there must be some $\tau$-open set $U$ with $f^{-1}(U)=A$. That is, there must be a $\tau$-open set that contains $a$ and not $b$.
So we have that $\tau$ is $T_1$. (Note: $\tau$ is also connected by the same argument, but that's not directly useful for the rest of this argument.)
We can now prove that $\tau$ refines the usual topology. Since $\tau$ is $T_1$, we know that $\mathbb R\setminus \{x\}$ is $\tau$-open for all $x$. For $a\in\mathbb{R}$, consider the function $f_a(x)=\max(a,x)$. This function is continuous and hence $\tau$-continuous. Since $\mathbb R\setminus \{a\}$ is $\tau$-open, its $f_a$-preimage must be $\tau$-open; that is, $(a,\infty)$ is $\tau$-open. We can prove $(-\infty,a)$ is open analogously. Intersecting open sets, we get that $(a, b)$ is $\tau$-open for every $a,b\in\mathbb{R}$.
If $τ$ is the weakest topology on $X$ such that $f:X→Y$ is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets under $Y$? This follows directly from the definition of a "continuous" function. Is this always the coarsest topology?
Yes. For $f$ to be continuous, you need the topology on $X$ to contain all preimages of open sets through $f$. The topology induced by a family $\mathcal T$ of functions is generated by
$$
\{f^{-1}(E):\ f\in\mathcal T,\ E\subset Y\ \text{ open }\}.
$$
Some reasons why one cares about these topologies are
They often appear naturally, as when considering duals and preduals of normed spaces;
In several cases the topology is coarse enough that some interesting sets become compact (for instance the unit ball in a Banach space, see the Banach-Alaoglu Theorem).
Best Answer
This follows from the connectedness of $(0,1)$. The sets $f^{-1}(0)$ and $f^{-1}(1)$ are open (by continuity) and clearly disjoint. If they were both non-empty, $(0,1)=f^{-1}(0)\cup f^{-1}(1)$ would be disconnected.