Please check the following proof.
Consider the function $f[a,b] \to \mathbb{R}$
Theorem :
$f[a,b] \to \mathbb{R}$ is continuous
$\implies$
$\forall \epsilon > 0$, $\exists \ \text{partition of}\ [a,b]$ $\ni (\forall\ \text{subinterval}) \ (\text{span } f < \epsilon)$
Proof:
$f[a,b] \to \mathbb{R}$ is continuous
$\implies$ $f$ is continuous at $a$
$\implies$ $\forall\ \epsilon>0$, $\left(\exists\ \delta \right) \left(-\delta < a-x < \delta\ \Rightarrow \dfrac{\epsilon}{2} < f(a)-f(x) < \dfrac{\epsilon}{2}\right)$
$\implies \left(\exists\ \delta_1\right) \left(-\delta_1 < a-x < \delta_1\ \Rightarrow -\dfrac{\epsilon_0}{2} < f(a)-f(x) < \dfrac{\epsilon_0}{2}\right)$
$\implies \left(\exists\ \delta_1\right) \left(x \in (a-\delta_1,a+\delta_1) \Rightarrow -\dfrac{\epsilon_0}{2} < f(a)-f(x) < \dfrac{\epsilon_0}{2}\right)$
$\implies$ $\left( -\dfrac{\epsilon_0}{2} < f(a)-f(x_{S1}) < \dfrac{\epsilon_0}{2} \right) \land \left( -\dfrac{\epsilon_0}{2} < f(x_{B1})-f(a) < \dfrac{\epsilon_0}{2} \right)$
$\implies$ $\left( 0 < f(a)-f(x_{S1}) < \dfrac{\epsilon_0}{2} \right) \land \left( 0 < f(x_{B1})-f(a) < \dfrac{\epsilon_0}{2} \right)$
$\implies$ $0 < f(x_{B1})-f(x_{S1}) < \epsilon_0$
$\implies$ span $f$ in $ (a-\delta_1,a+\delta 1 < \epsilon_0$
$\implies$ span $f$ in $ (a,a+\delta_1)=$ span $f$ in $ (a,a_1) < \epsilon_0$
Similarly:
span $f$ in $ (a_0,a_1) < \epsilon_0$
span $f$ in $ (a_1,a_2) < \epsilon_0$
span $f$ in $ (a_2,a_3) < \epsilon_0$
.....
span $f$ in $ (a_{(i-1)} ,a_i) < \epsilon_0$
span $f$ in $ (a_i,a_{(i+1)} ) < \epsilon_0$
where:
$a_i \in [a,b]$
$a_{(i+1)} \notin [a,b]$
Thus we see for partition $P_{[a,b]}=\{a=a_0, a_1, a_2,...,a_i, b \}$
$(\forall\ \text{subinterval}) \ (\text{span } f < \epsilon_0)$
Analogously $\forall\ \epsilon > 0$, we can make a partition of $[a,b] \ni (\forall\ \text{subinterval}) \ (\text{span } f < \epsilon)$
Please tell whether there are any mistakes in the above proof.
There is an issue with your proof. Why can you state that $b$ will be reached in a finite number of steps? "The $\delta_i$ may become smaller and smaller".
For example the map
$$\begin{array}{l|rcl} f : & [0,1) & \longrightarrow & \mathbb R \\ & x & \longmapsto & \frac{1}{1-x} \end{array}$$
is continuous but you won't be able to reach $1$ in a finite number of steps.
The compactness of $[a,b]$ has to come into play somewhere in proof of your theorem.