I believe that $\omega_1$ with the order topology is a counter example. Call a subset of $\omega_1$ *bounded* if it is contained in some initial interval $[0,\alpha]$ and unbounded otherwise.

**Lemma**: Any cover of $\omega_1$ by bounded open subsets is not minimal.

I am sure that this is standard, but here is a proof. It contains way too many notations and is a bit clumsy, I am afraid, but I hope it is readable.

We can assume that such a cover is is of cardinality $\omega_1$.
Let thus $\mathcal{V} = \{V_\alpha\,:\,\alpha\in\omega_1\}$ be a cover of $\omega_1$ by bounded open subsets.
By a standard argument the sets below are closed and unbounded in $\omega_1$:

\begin{align*} C_1 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\subset[0,\alpha)\}\\
C_2 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\supset[0,\alpha)\}
\end{align*}
Hence their intersection $C = \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta =[0,\alpha)\}$ is closed and unbounded as well. We may assume that each member of $C$ is a limit ordinal. When $\alpha\in \omega_1$,
let $\alpha^*$ denote the smallest element of $C$ which is $>\alpha$. For each $\alpha\in C$ choose $V_{s(\alpha)}\ni\alpha$ with $s(\alpha)\in[\alpha,\alpha^*)$. Since $V_{s(\alpha)}$ is open and $\alpha$ is limit, $V_{s(\alpha)}$ contains some interval $[\gamma(\alpha),\alpha]$. By Fodor's Lemma, there is some $\gamma\in\omega_1$ such that the set $S$ of $\alpha\in C$ such that $\gamma(\alpha) = \gamma$ is stationary and in particular unbounded. For any $\alpha\in S$ with $\alpha>\gamma^*$, $V_{s(\alpha)}$ contains in particular the interval $[\gamma^*,\alpha]$.
Fix now $\eta\in[\gamma^*,\gamma^{**})$. Choose $\alpha \ge \gamma^{**}$ in $S$. Then
$$ V_\eta \subset V_{s(\alpha)} \cup \left(\cup_{\beta<\gamma^*}V_\beta\right).
$$
Notice that $\gamma^*\le\eta<\alpha\le s(\alpha)$, hence the indices in the union at the righthand side do not contain $\eta$. It follows that $\cup (\mathcal{V} \backslash \{V_\eta\}) = \omega_1$.

**Corollary**: The cover of $\omega_1$ by the intervals $[0,\alpha]$, $\alpha\in\omega_1$, has no refinement that is a minimal cover.

countablecover has a minimal refinement. So the answer is yes, for $T_1$ Lindelof spaces. $\endgroup$ – Ramiro de la Vega Apr 26 '18 at 12:50