Assume $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$. Show that $f(1) \geq 2$ This question comes from Advanced Calculus, Fitzpatrick. Section 3.1 exercise 7

Suppose that the function $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $$f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$$
Show that $f(1) \geq 2$

Fitzpatrick uses the sequence definition of continuity, there is a similar question [here] (Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous and that $f\left(x\right)>2$)
I wanted to do this using epsilon-delta method. Is the following valid? Is there a simpler way to write this proof?
My attempt
First, assume $f(1) < 2$. Then there exists some $\varepsilon > 0$, call it $\varepsilon_0$, such that $2 - f(1) > \varepsilon_0$. Show this creates a contradiction.
Let $x_n = 1 - 1/n$ for all $\mathbb{N}$. Then $x_n$ is a sequence in $[0,1]$ that converges to 1. By definition, this is
$$\forall \delta_1 > 0 \; \exists N \in \mathbb{N}, \forall n \geq N \quad \vert x_n - 1 \vert < \delta_1$$
Since $f$ is continuous on $[0,1]$ it is continuous at $1$, therefore we have
$$\forall \varepsilon > 0 \; \exists \delta > 0 \; \forall x \in [0,1]\quad \vert x - 1 \vert < \delta \rightarrow \vert f(x) - f(1) \vert < \varepsilon$$
If we let $\varepsilon = \varepsilon_0$ and $\delta = \delta_1$. Then we know $\vert x_n - 1 \vert < \delta_1$, therefore we can conclude $\vert f(x_n) - f(1) \vert < \varepsilon_0$.
Furthermore, we know $\{f(x_n)\} >= 2$ for all $n$, so
$$\vert 2 - f(1) \vert \leq \vert f(x_n) - f(1) \vert < \varepsilon_0$$
$$\vert 2 - f(1) \vert < \varepsilon_0$$
$$-\varepsilon_0 < 2 - f(1) < \varepsilon_0$$
However this contradicts our assumption that $2 - f(1) > \varepsilon_0$
Therefore $f(1) >= 2$

You haven't exactly done this using the $\epsilon-\delta$ definition, though. Someone else has already commented on your proof so I'll just present my own argument.
Suppose that $f(1) < 2$. Since $f$ is continuous on $[0,1]$:
$$\lim_{x \to 1^-} f(x) = f(1) < 2$$
So, for each $\epsilon > 0$, there is a $\delta > 0$ such that:
$$1-x < \delta \implies |f(x)-f(1)| < \epsilon$$
Let $\epsilon = 2-f(1)$. Then, for a sufficiently small $\delta > 0$, we have:
$$1-x < \delta \implies f(1)-\epsilon < f(x) < f(1)+\epsilon$$
which implies that $f(x) < 2$ whenever $x \in (1-\delta,1]$. That's a contradiction. $\Box$
