Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 2

Proof
$f$ is not uniformly continuous.

Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:
 * $\size {x - y} < \delta$

and:
 * $\size {\map f x - \map f y} \ge \epsilon$

We will now choose for each $k \in \N$ numbers $x_k, y_k \in \closedint a b$ such that:
 * $\size {x_k - y_k} < \dfrac 1 k$

and:
 * $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$

We have that the sequence $\sequence {x_k}$ is bounded in $\closedint a b$.

So, by the Bolzano-Weierstrass Theorem there exists a convergent subsequence $\sequence {x_{k_j} }$ whose limit lies in $\closedint a b$.

Let this limit be denoted $x_0$.

Now we have:
 * $\size {x_0 - y_{k_j} } \le \size {x_0 - x_{k_j} } + \size {x_{k_j} - y_{k_j} } \le \size {x_0 - x_{k_j} } + \dfrac 1 {k_j}$

Therefore the sequence $\sequence {y_{k_j} }$ also converges to $x_0$.

Because $f$ is sequentially continuous at $x_0 \in \closedint a b$, we have:
 * $\ds \lim_{j \mathop \to \infty} \map f {x_{k_j} } = \map f {x_0} = \lim_{j \mathop \to \infty} \map f {y_{k_j} }$

This is however, a contradiction, as $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$ for all $k$ and thus all $k_j$.

Therefore $f$ is uniformly continuous.