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 $a,b\in\mathbb{R}$.
Let $X$ be topology on $\{1, 2, 3\}$ with base of $\{1\}, \{1, 2\}, \{1, 3\}$ and $Y$ be any Hausdorff space.
Assume $f(1) \neq f(2)$. Take open $U \subset Y$ s.t. $f(2) \in U$, $f(1) \notin U$. Then $2 \in f^{-1}(U)$, $1 \notin f^{-1}(U)$, so $f^{-1}(U)$ is not open. So $f(1) = f(2)$.
Analogously, $f(1) = f(3)$, and so $f$ is constant.
Best Answer
Yes, this is completely correct. Having fewer open sets in the codomain and/or more open sets in the domain leaves a continuous function still continuous in the new topology.