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 $a,b\in\mathbb{R}$.
Here is a positive proof, which encapsulates the same basic idea, but framing them in a different way.
Let $U_n$, for $n\in\Bbb N$ be any countable number of distinct open sets such that $x\in U_n$ for all $n$. Then for every $n$, there is some finite set $y_n$ such that $U_n=X\setminus y_n$. Therefore, by DeMorgan's law, $$X\setminus\bigcap U_n=\bigcup y_n\quad\text{and so}\quad \bigcap U_n=X\setminus\bigcup y_n.$$
But now $y_n$ are all finite sets, so we have a countable union of finite sets. Therefore this union is countable. So $X\setminus\bigcup y_n=\bigcap U_n$ is a co-countable set, which cannot be a singleton since $X$ is uncountable. So $\bigcap U_n\neq\{x\}$.
In particular, it is impossible that $x$ has a countable neighborhood basis. $\square$
The key point here is that when intersecting a countable number of open sets, the complement is a countable union of finite sets, and therefore cannot be equal to $X$ minus a single point.
Finally, your proof idea is flawed. For any $x\in\Bbb R$ consider $U_y=\Bbb R\setminus\{y\}$ as your open set for all $y\neq x$. These are all open sets such that $x\in U_y$. So there are uncountably many of them, since if $y\neq y'$ we have $y\in U_{y'}\setminus U_y$ and $y'\in U_y\setminus U_{y'}$. But nonetheless, $\Bbb R$ is first countable, and in fact second countable.
Best Answer
Your argument doesn't quite work, because the range could be a finite subset of $\mathbb R$ containing at least two elements. Then there is no guarantee that $f^{-1} (y)$ of the particular $y = f(x')$ that you have chosen is not co-finite. Importantly, $f$ may fail to be one-to-one, which appears to be something that you are assuming.
An important thing to remember about $\mathbb R$ with the usual topology is that it is Hausdorff. This is the central important fact for this problem. In fact, for any infinite space $X$ with the co-finite topology and any Hausdorff space $Y$, the only continuous functions $f : X \to Y$ are constant.
Proof. Suppose that $f : X \to Y$ is a nonconstant continuous function. Then there are distinct $u,v \in Y$ in the range of $f$. By Hausdorffness there are disjoint open sets $U,V \subseteq Y$ with $u \in U$ and $v \in V$. By continuity both $f^{-1}(U)$ and $f^{-1}(V)$ are open and nonempty, so they are co-finite. But then the intersection $f^{-1} (U) \cap f^{-1} (V)$ must be nonempty, which is impossible as $$f^{-1} (U) \cap f^{-1} (V) = f^{-1} ( U \cap V ) = f^{-1} ( \emptyset ) = \emptyset.$$