Suppose we have topology basis $\{(a,+\infty),a\in \mathbb R\}$ on $\mathbb R$. I wanted to find a continuous map from $[0,1]$, and $f(0)=a,f(1)=b$, where a and b are arbitrary real numbers.
From definition of continuous functions, we can find either closed sets and open sets in the topology and prove their preimage is the same. However, the open sets are only those in the basis, and the closed sets are their complements. While $f(1)=b$, then the preimage of $(-\infty,b]$ is some closed set $[m,1]$. How can a continuous function be constructed with such property?
Since every interval $(x,\infty)$ is open in $\Bbb R$ with its standard topology, the map $x\mapsto x$ from $\Bbb R$ with standard topology to $\Bbb R$ with your topology, is continuous. Likewise, $f\colon [0,1]\to\Bbb R$, $t\mapsto a+(b-a)t$ is continuous.