Alternative proof: Metric space is compact iff it's sequentially compact.

Question

Let $(X, d)$ be a metric space. The following are equivalent:

1. $X$ is compact.
2. $X$ is sequentially compact.

I saw two proofs of 2 ⇒ 1. One uses the Lebesgue number (in Munkres, theorem 28.2), and the other uses Lindelöf Property. I did it using the fact, that $(X,d)$ is compact iff all continuous functions $f: X → ℝ$ are bounded. It seems simpler. I would be glad for verification! Proof:

I will prove the contrapositive, ¬1 ⇒ ¬2. Assume that $X$ is not compact, even though it's sequentially compact.

Since $X$ is not compact there exists a continuous function $f:X→ℝ$ which is not bounded (WLOG, say by above). Thus there is a strictly increasing sequence $f(x_n)$ such that $n< f(x_n)$.

Since $X$ is sequentially compact, there exists a convergent subsequence of $x_n$, say $x_{n_k} → x$. Since $f$ is continous $f(x_{n_k}) → f(x)$. A contradiction, since only finitely many elements can be near $f(x)$, because after some $N$ for which $f(x) < N$, we have $N ≤ n_k ⇒ N ≤ f(x_{n_k})$.