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*Introduction to Real Analysis*by Bartle and Sherbert 3rd Edition, page 138, if anyone has access to it. The Theorem states:

Let I be a closed bounded interval and let f:I->R be continuous on I. Then f is uniformly continuous on I.

Proof(By Contradiction):

If f is not uniformly continuous on I then, by the preceding result(Nonuniform Continuity Criteria), there exists [tex]\epsilon_{0}>0[/tex] and two sequences [tex](x_n)[/tex] and [tex](u_n)[/tex] in I such that [tex]|x_n-u_n|<1/n[/tex] and [tex]|f(x_n)-f(u_n)|\geq\epsilon_{0}[/tex] for all [tex]n\in\mathbb{N}[/tex]. Since I is bounded, the sequence [tex](x_n)[/tex] is bounded; by the Bolzano-Weierstrass Theorem there is a subsequence [tex](x_n_k)[/tex] of [tex](x_n)[/tex] that converges to an element z. Since I is closed, the limit z belongs to I. It is clear that the corresponding subsequence [tex](u_n_k)[/tex] also converges to z, since

[tex]|u_n_k - z|\leq|u_n_k - x_n_k| + |x_n_k - z|[/tex].Now if f is continuous at the point z, then both of the sequences [tex](f(x_n_k))[/tex] and [tex](f(u_n_k))[/tex] must converge to [tex[f(z)[/tex]. But this is not possible since

[tex]|f(x_n)-f(u_n)|\geq\epsilon_0[/tex]

I understand the proof up to the part where it says it is clear that [tex](u_n_k)[/tex] also converges to z. I don't know why showing this inequality shows that the subsequence converges to z. Also why is the difference between the two sequences less then 1/n?