Question: Consider the function $f(x)=[x]$ on the interval $[0,2]$ where $[x]$ denotes the largest integer less than or equal to x.

Is this function continuous? I cant find a reason for it not to be, although im not sure.

Answer: 
The function $\lfloor x \rfloor$ is not continuous at interior points that are integers. In this case it at $x=1$. It is also discontinuous at $x=2$

If you are looking for a formal argument here is one.

If a function $f(x)$ is continuous at $x=a$ then for every $\varepsilon>0$ there exist an $\delta>0$ such that $$|x-a|<\delta \Rightarrow |f(x)-f(a)| < \varepsilon $$

So if the function is continuous at $x=1$ then for $\varepsilon=\frac12$ there should be a $\delta$. If there is a delta at points such that $1-\delta<x<1+\delta$ the inequality $|f(x)-1| < \frac12 \, \, \star$ should be satisfied.

Now our job is reduced to show that whatever $\delta$ we choose there is a point for which $\star$ do not hold.

So if we choose a $\delta < 1$ the point $0<1-\frac\delta2<1$ and $f(x)=0$ at all such regions implying that $|f(x)-1|=|0-1|=1>\frac12$.
It is not continous.

To see that, consider for example that $f(1) = 1$, but $f(1 - \epsilon) = 0$ for any $\epsilon>0$. There is a 'jump' from $f=0$ to $f=1$ without any values inbetween, regardless of the size of $\epsilon$.