Let $X,Y$ be topological space and $Y$ locally compact. Define $\alpha$ relation on $X$. Consider identity relation on $Y$.(i.e. $Y/1=Y$.)
Let $T\subset X\times Y/(\alpha\times 1)$ be open. Then $T'$ preimage of $T$ in $X\times Y$ is product of open sets $U\times V$ with open sets $U\subset X,V\subset Y$. Then the book claims we can assume $V$ to be compact set.
$\textbf{Q:}$ How can one assume $V$ to be compact? Normally, I knew there is a $V'$ smaller neighborhood of $V$ contained in $K$ with $K$ compact by $Y$ locally compact. However, I need to trim down this $\bar{V'}\subset V$ to reduce $K$ to $\bar{V'}$. This is achieved by $Y$ hausdorff property.
Ref. Pg 2 Algebraic Topology Switzer
It suffices to demonstrate 2 definitions of local compactness are equivalent under assumption of hausdorff property.
(a) Every open neighbourhood has a compact neighborhood(i.e. $\forall U,\exists K\subset U$ with $U$ open and $K$ compact.)
(b) For every point there is a compact set containing an open set.
Clearly $a)\implies b)$. It suffices to deal with converse implication.
Take $U\subset K$. It suffices to produce $\bar{W}\subset U$ with $\bar{W}$ compact. Now consider $\partial U\subset K$ is compact by $K$ compact. Use hausdorff property to find a open set $V$ disjoint from $\partial U$'s covering and containing a point $x\in U$. Here we need to use compactness of $\partial U$ to ensure finite intersection. Now $W=U\cap V$ is an open set containing $x$. Note that $\bar{W}\neq K$ for sure as we have removed $\partial U$. So we have found $\bar{W}$ compact neighborhood contained in $U$.
The general case can be handled similarly.