Uniformly Continuous Function is Continuous/Real Function/Proof 2

Theorem
Let $I$ be an interval of $\R$.

Let $f: I \to \R$ be a uniformly continuous real function on $I$.

Then $f$ is continuous on $I$.

Proof
Let $x \in I$.

Let $\epsilon \in \R_{>0}$.

As $f$ is uniformly continuous:
 * $\exists \delta \in \R_{>0}: \left({x, y \in I, \left\vert{x - y}\right\vert < \delta \implies \left\vert{f \left({x}\right) - f \left({y}\right)} \right\vert < \epsilon}\right)$

Then, for any $y \in I$ such that $\left\vert{x - y}\right\vert < \delta$:
 * $\left\vert{f \left({x}\right) - f \left({y}\right)} \right\vert < \epsilon$

Thus by definition $f$ is continuous at $x$.

As $x$ was arbitrary, $f$ is continuous on all of $I$.