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 $|f(x)-1|=|0-1|=1>\frac12$.
If you're familiar with the properties of continuous functions, another way to see it is that $f^{-1}(1)=[0,1)$, which is not a closed subset of $[0,2]$.