Question: How do we do this? Does it involve the fact that we know $f$ is continuous, and therefore the limit approaching the endpoint $1$ should be the same coming from both sides?

Answer: 
If $f(1)<2$ then $\epsilon=2-f(1)$ is positive. Since $f$ is continuous and $f(0)\ge 2$, then there exists some $c\in[0,1)$ such that $f(c)=2-\frac\epsilon2$. Contradiction.
An alternative topological solution:

Since $f$ is continuous, $f(\overline{A}) \subset \overline{f(A)}$ for all $A \subset [0,1]$

Applying this to $[0,1)$:

$f([0,1]) \subset \overline{f([0,1))} \subset \overline{[2,\infty)} = [2,\infty)$