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Let $f : [a, b] \to \mathbb{R}$ be continuous on $[a, b]$. Suppose that for each $n \in \mathbb{N}$ there is a point $x_n ∈ [a, b]$ such that $|f(x_n) − \alpha| < 1/n$. Use the Bolzano–Weierstrass Theorem to show that there is a point $x^* ∈ [a, b]$ such that $f(x^*) = \alpha$.

user1337
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Let $y$ be an accumulation point of $x_n$. Therefore, there is a subsequence $x_{n_k}$ that $x_{n_k}\to y$. Then $0\leq|f(x_{n_k})-a|<1/n_k$. Taking limits, and using that $f$ is continuous you get that $0\leq |f(y)-a|\leq 0$

Therefore, $f(y)=a$.