On page 97 of Kreyzig's functional analysis book he provides a proof that a linear operator $T$ is continuous if and only if it is bounded.
When proving that $T$ is continuous implies that $T$ is bounded he says that if we assume $T$ is continuous at some $x_0$ then for any $\varepsilon >0$ there exists a $\delta > 0$ such that $\Vert Tx - Tx_0 \Vert \le \varepsilon$ for all $x$ satisfying $\Vert x - x_0 \Vert \le \delta$.
But on the previous page where he gave the definition of a continuous operator $T$ he used strictly less inequalities for $\varepsilon$ and $\delta$. That is, $T$ is continuous at some $x_0$ means that for any $\varepsilon >0$ there exists a $\delta > 0$ such that $\Vert Tx - Tx_0 \Vert < \varepsilon$ for all $x$ satisfying $\Vert x - x_0 \Vert < \delta$.
How can he extend the strict inequality to a less than or equal to inequality?
If you know that you can show this with $\le$ and need $<$ just apply the $\le$ case to $\frac{\varepsilon}{2}$ in order to get the $<$ for $\varepsilon$ (and vice versa)