Suppose first that $X$ is not countably compact, and let $\mathscr{U}=\{U_n:n\in\omega\}$ be a countable open cover of $X$ with no finite subcover. For $n\in\omega$ let $V_n=\bigcup\limits_{k\le n}U_k$, and let $\mathscr{V}=\{V_n:n\in\omega\}$. Clearly $\mathscr{V}$ has no finite subcover. Suppose that $A\subseteq \omega$ is infinite, and let $m=\min A$; then every point of $V_m$ is in every member of $\{V_n:n\in A\}$, which is therefore not even point-finite, let alone locally finite. It follows that $X$ must be countably compact.
On the other hand, $X$ is obviously paracompact and hence metacompact, and it’s a standard result that every metacompact countably compact space is compact.
Thus, the hypothesis is actually stronger than necessary: it’s sufficient to assume that every open cover has a point-finite subcover.
Added: The proof isn’t very hard. Let $\mathscr{U}$ be an open cover of $X$; by hypothesis $\mathscr{U}$ has a point-finite subcover $\mathscr{V}$. Let $\mathfrak{S}$ be the family of subcovers of $\mathscr{V}$, ordered by $\supseteq$. Suppose that $\mathfrak{C}$ is a chain in $\mathfrak{S}$, and let $\mathscr{C}=\bigcap\mathfrak{C}$; clearly $\mathscr{C}\subseteq\mathscr{V}$. If $\mathscr{C}$ isn’t a cover of $X$, let $x\in X\setminus\bigcup\mathscr{C}$. Let $\mathscr{V}_x=\{V\in\mathscr{V}:x\in V\,\}$; for each $V\in\mathscr{V}_x$ there is a $\mathscr{C}_V\in\mathfrak{C}$ such that $V\notin\mathscr{C}_V$. $\mathscr{V}_x$ is finite, so we can enumerate $\mathscr{V}_x=\{V_1,\dots,V_n\}$ in such a way that $\mathscr{C}_{V_1}\supseteq\dots\supseteq \mathscr{C}_{V_n}$. But then $x\notin\bigcup\mathscr{C}_{V_n}$, which is impossible, since $\mathscr{C}_{V_n}$ is a cover of $X$. Thus, $\mathscr{C}$ does cover $X$, so $\mathscr{C}\in\mathfrak{S}$ and is clearly an upper bound for $\mathfrak{C}$ in $\mathfrak{S}$. $\mathfrak{C}$ was an arbitrary chain in $\mathfrak{S}$, so by Zorn’s lemma $\mathfrak{S}$ has a $\supseteq$-maximal element $\mathscr{W}$. In other words, $\mathscr{W}$ is a subcover of $\mathscr{V}$ with no proper subcover: every member of $\mathscr{W}$ contains at least one point of $X$ that is not in any other member of $\mathscr{W}$. Such a cover is said to be irreducible, and we’ve just shown that every point-finite open cover has an irreducible subcover.
Finally, observe that an irreducible open cover $\mathscr{W}$ of a countably compact space must be finite. If not, let $\mathscr{W}_0=\{W_n:n\in\omega\}$ be an infinite subset of $\mathscr{W}$, and let $W=\bigcup\Big(\mathscr{W}\setminus\mathscr{W}_0\Big)$; then $\mathscr{W}_0\cup\{W\}$ is a countable open cover of $X$ with no finite subcover, since each of the sets $W_n$ contains at least one point not covered by any other member of the cover.