Here’s a counterexample to the conjecture.
Let $Y=\Bbb N\times\Bbb N$, let $p$ and $q$ be distinct points not in $Y$, and let $X=Y\cup\{p,q\}$. Points in $Y$ are isolated. For each $k\in\Bbb N$ the set
$$B_p(k)=\{p\}\cup\left\{\langle m,n\rangle\in Y:n\ge k\right\}$$
is a basic open nbhd of $p$. For each $k\in\Bbb N$ the set
$$B_q(k)=\{q\}\cup\left\{\langle m,n\rangle\in Y:m\ge k\right\}$$
is a basic open nbhd of $q$. For convenience, for $n\in\Bbb N$ let $S_n=\{n\}\times\Bbb N$.
Each open nbhd of $p$ contains all but finitely many points of each $S_n$, and each open nbhd of $q$ contains all but finitely many of the sets $S_n$. Thus, every open set containing both $p$ and $q$ contains all but finitely many points of $Y$, and $X$ is compact. $Y$ is Hausdorff, as are the sets $B_p(k)$ and $B_q(k)$, so $X$ is locally Hausdorff. However, none of the sets $B_p(k)$ contains a compact nbhd of $p$, so $X$ is not locally compact. (Similarly, $X$ fails to be locally compact at $q$.)
Overall the hypotheses are gross overkill, but you can’t omit any of them. Matt Samuel has already given the example of the long line to show that you can’t omit second countability.
The particular point topology on a countably infinite set $X$ is non-Hausdorff, locally compact, and second countable — if $p$ is the particular point, $\mathscr{B}=\big\{\{p,x\}:x\in X\big\}$ is a countable base for the topology — and it’s not paracompact, since the open cover $\mathscr{B}$ has no open refinement that is locally finite at $p$. This shows that you can’t omit Hausdorffness.
To see that you can’t omit local compactness, let
$$\begin{align*}
D&=\left\{\left\langle\frac1m,\frac1n\right\rangle:m,n\in\Bbb Z^+\right\}\;,\\\\
H&=\left\{\left\langle\frac1m,0\right\rangle:m\in\Bbb Z^+\right\}\;,\text{ and}\\\\
X&=\{\langle 0,0\rangle\}\cup H\cup D\;.
\end{align*}$$
Points of $D$ are isolated. For $m,n\in\Bbb Z^+$ let $x_m=\left\langle\frac1m,0\right\rangle$ and $y_{m,n}=\left\langle\frac1m,\frac1n\right\rangle$, and define
$$B_n(x_m)=\{x_m\}\cup\{y_{m,k}:k\ge n\}\;;$$
$\{B_n(x_m):n\in\Bbb Z^+\}$ is a local base at $x_m$. (In other words, the sequence $\langle y_{m,n}:n\in\Bbb Z^+\rangle$ converges to $x_m$.) Let $p=\langle 0,0\rangle$, and for $n\in\Bbb Z^+$ let
$$B_m(p)=\{y_{k,n}:k\ge m\}\;;$$
$\{B_m(p):m\in\Bbb Z^+\}$ is a local base at $p$. With this topology $X$ is Hausdorff and second countable, but it’s not locally compact at $p$. It’s also not paracompact: the open cover
$$\mathscr{U}=\{B_1(p)\}\cup\{B_1(x_n):n\in\Bbb Z^+\}$$
has no open refinement that is locally finite at $p$.
The only real function served by local compactness is to make the space regular: every locally compact Hausdorff space is regular. If we require the space to be regular and Hausdorff, we can drop the assumption of local compactness, because every second countable space is Lindelöf, and every $T_3$ Lindelöf space is paracompact. Second countability is of course a good deal stronger than the Lindelöf property, so this theorem is really just a weak version of the theorem that a $T_3$ Lindelöf space is paracompact.
Best Answer
It is not even true that compact Hausdorff spaces are second-countable. (Of course, a compact Hausdorff space is locally compact and can be covered by a countable family of compact subsets.)
An example is the ordinal space $[ 0 , \omega_1 ]$ (where $\omega_1$ denotes the least uncountable ordinal).
More information about this space can be found on the following post on Dan Ma's Topology Blog:
(If you want a non-compact example of a locally compact σ-compact Hausdorff space that is not second-countable you can just take a disjoint union of countably-infinite copies of $[0,\omega_1]$.)