There are several problems with the argument. First, it isn’t always possible to find an irreducible subcover. (That’s the standard name for what you’re calling a maximal subcover — and what you really mean is that it’s a minimal subcover.)
Example. For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$; $\{\Bbb N\}\cup\{U_n:n\in\Bbb N\}$ is a topology on $\Bbb N$. Let $\mathscr{U}=\{U_n:n\in\Bbb N\}$; $\mathscr{U}$ is an open cover of $X$. Suppose that $\mathscr{V}\subseteq\mathscr{U}$; $\mathscr{V}$ covers the space if and only if $\{n\in\Bbb N:U_n\in\mathscr{V}\}$ is infinite. Thus, if $\mathscr{V}$ is any subcover of $\mathscr{U}$, we can remove any finite number of members of $\mathscr{V}$ and still have a cover of the space.
Fortunately, it turns out that we don’t actually need to get one.
Secondly, in order to get a sequence by your approach, you need the subcover to be countable. We can ensure that by starting with a countable open cover.
Suppose that $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a countable open cover of $X$ having no finite subcover. For $n\in\Bbb N$ let $V_n=\bigcup_{k\le n}U_k$, and let $\mathscr{V}=\{V_n:n\in\Bbb N\}$; clearly $\mathscr{V}$ is an open cover of $X$ with no finite subcover, and $V_0\subseteq V_1\subseteq V_2\subseteq\ldots\;$. Let $n_0=0$, and choose any $x_0\in V_0$. $V_0\ne X$, so there is an $n_1>0$ such that $V_0\subsetneqq V_{n_1}$; choose any $x_1\in V_{n_1}\setminus V_0$. Suppose that $n\in\Bbb N$, and we’ve chosen $n_k$ and $x_k$ for $k=0,\ldots,n$ so that $x_k\in V_{n_k}\setminus V_{n_{k-1}}$ for $k=1,\ldots,n$. $V_{n_k}\ne X$, so there is an $n_{k+1}>n_k$ such that $V_{n_k}\subsetneqq V_{n_{k+1}}$, and we choose $x_{n+1}\in V_{n_{k+1}}\setminus V_{n_k}$. In this way we construct an infinite sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points of $X$. Let $p$ be an accumulation point of this sequence. $\{V_{n_k}:k\in\Bbb N\}$ covers $X$ (why?), so there is an $m\in\Bbb N$ such that $p\in V_{n_m}$. But then $V_{n_m}$ is an open nbhd of $p$ that does not contain any of the points $x_k$ for $k>m$, contradicting the assumption that $p$ is an accumulation point of the sequence.
I’m assuming that when you say that $p$ is an accumulation point of a sequence $\langle x_n:n\in\Bbb N\rangle$, you mean that each open nbhd of $p$ contains infinitely many of the terms $x_n$. If you simply mean that $p$ is an accumulation point of the set $\{x_n:n\in\Bbb N\}$, then you need to assume that $X$ is $T_1$ in order to get your contradiction, since you need to know that $p$ has an open nbhd disjoint from $\{x_k:k\le m\}\setminus\{p\}$ as well.
The actual theorem here is that if every sequence in $X$ has an accumulation point, then $X$ is countably compact: every countable open cover has a finite subcover. Virtually the same proof shows that if $X$ is $T_1$, and every infinite set in $X$ has an accumulation point, then $X$ is countably compact.
