Yes! Since one have that
$$
(a,b) = \cup_{n\ge 1} \ [a+\frac{\epsilon}{n},b)
$$
where $\epsilon < \frac{b-a}{2}$.
Note that if for topology ${\mathcal T}_1$ with basis ${\mathcal S}_1$ and topology ${\mathcal T}_2$, one have that ${\mathcal S}_1 \subseteq {\mathcal T}_2$, then ${\mathcal T}_2$ is finer than ${\mathcal T}_1$. In this case if ${\mathcal S}_2$ be a basis for topology ${\mathcal T}_2$ and ${\mathcal S}_2 \not\subseteq{\mathcal T}_1$, then ${\mathcal T}_2$ is strictly finer than ${\mathcal T}_1$.For this, it is enough to note that $[0,1)$ is not open in standard topology.
For more details you can consult this textbook: James Munkres, "Topology; A First Course".
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 $a,b\in\mathbb{R}$.
Best Answer
It's the identity function he discusses, i.e. $f(x)=x$, which is a well-defined function from $\Bbb R$ to $\Bbb R_l$ because both spaces have the same elements (all real numbers). And of course $a \in \Bbb R$. Maybe the arbitrary $a$ and $b$ confuse you? To show $f$ is not continuous it is enough to find one open set in $\Bbb R_l$ such that $f^{-1}[O]$ is not open in $\Bbb R$. I'll take $O=[0,1)$ for definiteness, which is open in $\Bbb R_l$ by definition.
Recall that $f^{-1}[[0,1)]= \{x \in \Bbb R \mid f(x) \in [0,1)\} = \{x \in \Bbb R \mid x \in [0,1)\} = [0,1)$ and $[0,1)$ is not open in $\Bbb R$ as no open interval that contains $0$ sits inside $[0,1)$ so that $0$ is not an interior point of $[0,1)$ in $\Bbb R$.
So we are done showing $f$ non-continuous. Note that inverse images of the identity function are very simple: $f^{-1}[O]=O$ for all $O$. And also $f[O]=O$.