Continuity and lower bound I was initially presented with the following problem:
Show that $x < \tan{x}$ for all $x \in (0, \pi/2)$.
My solution involved setting $f(x) = \tan{x}-x$ and then showing that $f(x)$ is strictly increasing on $(0, \pi/2)$. However, I need the following lemma proved to complete (I know that there is an alternative solution with the Mean Value Theorem, but I'm interested in whether or not my proof for the lemma is correct):
Lemma: If $f$ is continuous on $[a, b)$ and increasing on $(a,b)$, $f(a) \le f(x)$ for all $x \in (a, b)$.
My proof:
EDIT: I realize this proof is incorrect. Is the lemma correct? What other proof can I use?
Assume that there exists a $x_0 \in (a, b)$ such that $f(a) > f(x_0)$.
Let $\delta = x_0 - a$. Note that $\delta > 0$ since $x_0 > a$.
We may construct a sequence $(x_n)$ such that $x_n = a + \frac{\delta}{n}$.
Since $x_n < x_0$ for all $n \in \mathbb{N}$ and $f$ is increasing on $(a, b)$, $f(x_n) < f(x_0) < f(a)$ for all $n \in \mathbb{N}$. 
Notice that 
$\lim{x_n} = 0$
However, since $f$ is continous on $[a, b)$, 
$\lim f(x_n) = f(a)$
which is a contradiction since $f(x_n) < f(a)$ for all $n \in \mathbb{N}$.
Is that a rigorous proof of the lemma presented?
Thanks

First let me give an insight to the proof. I am trying to prove that if there is an $x_0$ such that $f(a)>f(x_0)$ then under those circumstances $f$ cannot be continuous at $a$.

As $f$ is increasing on $\left(a,b\right)$, hence note that for every $x_1,x_0\in (a,b)$ and $x_1 < x_0 \implies f\left(x_1\right)\le f\left(x_0\right)$. 
Now if there is an $x_0$ such that $f(x_0)<f(a)$ using the above idea for every $x<a\implies f(x)\le f(x_0)<f(a)$$\implies 0<f(a)-f(x_0)\le f(a) - f(x) \quad\left(\color{blue}{\star}\right)$
As $f$ is continuous at $a$ for $\large \varepsilon =\frac{f(a)-f(x_0)}{2}$ there exists a $\delta$ such that $\large x<a+\delta\implies f(a)-f(x)<\frac{f(a)-f(x_0)}{2}$ which is a contradiction to $\left(\color{blue}{\star}\right)$.
