Is the following a valid proof that $X$, an uncountable set, with the cofinite topology is not first-countable?
By way of contradiction, suppose that $\mathscr{B}_x=\left\{B_n:n\in\mathbb{N}\right\}$ is a countable local base of open sets at the point $x$.
Since $X$ is $T_1$ and $\mathscr{B}_x$ is a local base at $x$,
$$\bigcap_{n\in\mathbb{N}} B_n = \{x\}.$$
By De Morgan,
$$X\setminus\bigcap_{n\in\mathbb{N}} B_n = X\setminus\{x\}$$
$$\bigcup_{n\in\mathbb{N}} X\setminus B_n = X\setminus\{x\}.$$
As $B_n$ is open the set $X\setminus B_n$ is finite, hence, it is countable. Notice, we have a countable union of countable sets, hence, a countable set. Therefore, it must be that $X\setminus\{x\}$ is countable, which implies $X$ is countable, a contradiction.
Best Answer
Using $\bigcap_n B_n = \{x\}$ is not directly necessary, but is OK in your proof. So your solution is fine as soon as you can justify this intersection from $T_1$-ness (some earlier result in your text presumably).
You could also argue: $\bigcup_n X\setminus B_n$ is at most countable, so we can pick $y \in X$ that is not in this union and unqual to $x$, as $X\setminus \{x\}$ is uncountable. Then $U= X\setminus \{y\}$ is open (definition of cofinite topology, it has finite complement) contains $x$ and so must contain some $B_m$. But then $y$ was chosen to be not in $X\setminus B_m$, so $y \in B_m$ and we cannot have $B_m \subseteq X\setminus \{y\}$, contradiction.