Edit: My original post contained a faulty counterexample with a rather obvious flaw. Since the rest of the post contained some useful (but well-known) material, I will leave this up as an extended comment.
So you want to consider a space $X$ which is the colimit of an expanding sequence
$$X_1\subseteq X_2\subseteq\dots\subseteq X_n\subseteq\dots$$
of its compact (not necessarily Hausdorff) subspaces? Such an $X$ is Lindelöf, and is $T_1$ when each $X_n$ is $T_1$.
In case $X$ is $T_1$, it is hemicompact.
Proof: Suppose that $K\subseteq X$ is a compact subset not contained in any $X_n$. We can assume without loss of generality that the $X_n$ are all distinct and that $K\cap X_1\neq\emptyset$. For each $n\geq0$ choose $x_n\in K\cap(X_n\setminus X_{n-1})$, where we understand $X_{0}=\emptyset$. Write $D=\{x_n\mid n\in\mathbb{N}\}\subseteq X$. If $C\subseteq D$ is any subset, then $C$ is closed in $D$, since for each $n$ we have that $C\cap X_n$ is finite and hence closed in $X_n$. It follows that $D$ is infinite, discrete and closed in $X$, and this contradicts the compactness of $K$. $\;\blacksquare$
On the other hand, with not separation assumptions we have;
If $X_n$ is open in $X_{n+1}$ for each $n\in\mathbb{N}$, then $X$ is hemicompact.
Proof: The family $\{X_n\}_{n\in\mathbb{N}}$ is a an open cover of $X$. $\;\blacksquare$
Example 0: A hemicompact example which is not exhaustible by compacts.
For $n\in\mathbb{N}$ let $X_n=\bigvee^n_{i=1}S^1$ and equip $X=\bigcup^\infty_{i=1}X_n$ with the colimit topology. As a countable CW complex, $X$ is a Hausdorff $k_\omega$-space. If $X$ had a sequence of exhaustion $K_1\subseteq int(K_2)\subseteq K_2\dots$, then each point of $X$ would be contained in the interior of one of the compact $K_n$'s. Since each compact subset of a CW complex is metrisable, each point of $X$ would therefore have a metrisable neighourhood. But a locally metrisable paracompact space is metrisable, and $X$ is not even first-countable at the wedge point. $\;\square$
The rational numbers $\mathbb Q$ are $\sigma$-compact but not hemicompact. The ordinal space $Y=\omega_1+1=[0,\omega_1]$ is compact, but not second countable (it's not first countable at $\omega_1$ for example).
So their topological sum $X=\mathbb Q\coprod Y$ is $\sigma$-compact, but not second countable (because of $Y$), and not hemicompact (because of its closed subspace $\mathbb Q$).
Best Answer
A very simple example is the rational numbers. Obviously $\mathbb{Q}$ is $\sigma$-compact by covering it with singletons. However, I claim it is not hemicompact. Indeed, suppose $(K_n)$ is a sequence of compact subsets of $\mathbb{Q}$. Each $K_n$ does not contain any neighborhood of $0$, so we can pick a sequence $(x_n)$ where each $x_n\not\in K_n$ and $x_n\to 0$. Then the set $\{x_n:n\in\mathbb{N}\}\cup\{0\}$ is compact but not contained in any $K_n$.
More generally, this argument shows that a hemicompact first-countable space must be locally compact (in the weak sense of having a compact neighborhood of every point). So any $\sigma$-compact first-countable space that is not locally compact is a counterexample.