= Solution
Include the empty set, and take all unions of intervals $[a,b)$ with $a<b$. These intervals cover the real line, and the intersection of two is either empty or $[\max(a,c),\min(b,d))$. Distributing intersections over unions therefore makes finite intersections of open sets open; arbitrary unions are open by construction. The whole line is open, so the axioms of a <topology> hold. This is the <lower limit topology>, whose space is the <Sorgenfrey line>.
If $x<y$, the neighborhoods $[x,y)$ and $[y,y+1)$ are disjoint. Hence $\boxed{(\mathbb R,\tau_1)\text{ is Hausdorff}}$.
For the identity into the <cofinite topology>, an inverse image is the same subset of $\mathbb R$. Let $F$ be finite and $x\notin F$. If $F$ has a point greater than $x$, choose $b$ to be its least such point; otherwise take $b=x+1$. Then $[x,b)$ avoids every point of $F$, including $b$ itself. Taking these intervals over all $x\notin F$ writes $\mathbb R\setminus F$ as a $\tau_1$-open set. The inverse image of every cofinite open set, and of the empty set, is open. \b[The identity map is continuous], because $\tau_2\subseteq\tau_1$.
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