Another proof: let $\mathcal{B} = \{B_i \mid i \in I \}$ be any base for $X$ and $\mathcal{C} = \{C_n \mid n \in \mathbb{N} \}$ be a countable base for $X$. Consider the set of pairs $$\mathcal{P} = \{ (n,m) \in \mathbb{N} \times \mathbb{N} \mid \exists i \in I: C_n \subset B_i \subset C_m \, \}$$
which is countable, as a subset of all pairs from a countable set. For each $(n,m) \in \mathcal{P}$ fix $i = i(n,m)$ as the $i$ in the definition of $\mathcal{P}$. The set $\mathcal{B}' = \{ B_{i(n,m)} \mid (n,m) \in \mathcal{P} \,\}$ is thus a countable subfamily of $\mathcal{B}$ and is the required base:
Let $O$ be any open subset of $X$ and let $x \in O$. As $\mathcal{C}$ is a base, there exists some $m \in \mathbb{N}$ such that $x \in C_m \subset O$, and applying $\mathcal{B}$ is a base we find $j \in I$ such that $x \in B_j \subset C_m$, and again applying that $\mathcal{C}$ is a base, we find $C_n$ so that $x \in C_n \subset B_j$.
Now note that $(n,m)$ is in $\mathcal{P}$ as witnessed by $j$. We don't know that $j = i(n,m)$ but we don't care, because for $i = i(n,m)$ we also have $x \in B_i \subset O$, as required.
So we have found a member of $\mathcal{B}'$ between every open $O$ and its elements, showing $\mathcal{B}'$ is a base.
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.
Best Answer
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 $\mathbb Q$).