Compare definitions of locally compact spaces I already searched and saw some questions about definitions of locally compact spaces (like Two definitions of locally compact space), but I could NOT find the exact comparison of the two definitions I want to discuss below:
Definition 1: Let T=(S,τ) be a topological space. Then T is locally compact iff every point of S has a local basis $\mathscr B$ such that all elements of $\mathscr B$ are compact. (from https://proofwiki.org/wiki/Definition:Locally_Compact)
Definition 2: Let T=(S,τ) be a topological space. Then T is locally compact iff - a) T is Hausdorff; and b) Every point has a compact neighbourhood (from https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf)
I could see that from Definition 1, it is easy to get property b of Definition 2
From https://proofwiki.org/wiki/Definition:Locally_Compact, it comments that "if T is a Hausdorff space, then Definition 1 and Definition 2 are equivalent" - could anyone help to prove the direction from Definition 2 to Definition 1 when T is a Hausdorff space?
 A: For some point $x$ take its compact neighborhood $K$. 
We want to show that for any open neighborhood $U$ of $x$, there is some compact neighborhood $K'\subseteq U$ and we will use $K$ to construct it.
So, take any open neighborhood $U$ of $x$. If $K\subseteq U$, we are done. Otherwise, $K\cap U^c$ is closed subset of $K$, thus compact. Since space is Hausdorff, we can find open $V$ and $V'$ such that $K\cap U^c\subseteq V$ and $x\in V'$, $V\cap V'=\emptyset$. Define $K' = K\cap V^c \subseteq K$. You can easily check that $K'$ is compact neighborhood of $x$ contained in $U$.
Edit:
Let me clarify, in Hausdorff space you can separate compact sets from points. Let $K$ be compact and $x\not\in K$. For each $y\in K$ choose open $y\in U_y$, $x\in V_y$ such that $U_y\cap V_y=\emptyset$. $\{U_y\mid y\in K\}$ is open cover for $K$, so we can find finite subcover $\{U_{y_1},\ldots,U_{y_n}\}$ of $K$. Define $U=\cup_{i=1}^n U_{y_i}$, $V=\cap_{i=1}^n V_{y_i}$. Then $K\subseteq U$, $x\in V$, $U\cap V =\emptyset$.
A: To get from Def'n 2 to Def'n 1 for a Hausdorff space. For  $p\in S$ let $U$ be a compact nbhd of $p.$ Let $V\in \tau$ with $p\in V\subset U.$ Then $Cl_X(V)$ is compact, because it  is a closed subset of the compact Hausdorff space $U.$
(i). Observe that for all $W\subset V$ we have ($W$ is open in the space $U$) iff $W\in \tau.$
Let $B$ be an open local base (basis) at $p$ in the space $X.$ Then $C=\{V\cap b:b\in B\}$  is an open local base at $p$ in the space  $X$ and also in the space  $U$, by (i).
Now a compact Hausdorff space is a regular space. So $U$ is regular.
So for each $c\in C$ we can choose $c'$ where $c'$ is open in $U$ and $p\in c'\subset Cl_U(c')\subset c.$  Let $C'=\{c': c\in C\}.$  By (i), $C'$ is a local open base at $p.$ And $Cl_U(c')$ (which is equal to $Cl_X(c'))$ is a closed subset of the compact Hausdorff space space $U$ so $Cl_U(c')$ is compact.
Now for $p\in Y\in T,$ take  $c\in C$ with $p\in c \subset Y.$ Then  $\;\overline {c'}\;$ is a  compact nbhd  of $p,$ and is a subset of $Y. $ And we are done.
NOTE: To show that a compact Hausdorff $S$ space is regular, let $p\in S$ with $p\not \in A=\bar A\subset S.$ To find disjoint open $U,V$  with $p\in U$ and $A\subset V$: 
(i). If $A$ is empty let $U=S$ and $V=\emptyset.$
(ii). If $A$ is not empty then for each $a\in A$ let $U_a,V_a$ be disjoint open sets with $p\in U_a$ and $a\in B_a.$ Now $A$ is compact (as it is a closed subset of the compact Hausdorff space $S$) so the open cover $\{B_a:a\in A\}$ of $A$ has a finite non-empty sub-cover $\{B_a:a\in F\}$ where $F$ is a finite subset of $A.$ 
Let $U=\cap_{a\in F}U_a$  and $V=\cup_{a\in F}B_a.$
