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: 
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 $\epsilon$.
Another characterisation of a function $f$ being continous at an argument value $x$ is, that it has a value $f(x)$ there which is equal to its limit value there. This requires an existing limit value. 

That last requirement is not valid for your function at $x = 1$:

If you arrive from the left, the limit at $x = 1$ is $0$,
$$ 
\lim\_{x \to 1 \atop x < 1} \left[ x \right] = 
\lim\_{x \to 1 \atop 0 < x < 1} \left[ x \right] = 
\lim\_{x \to 1 \atop 0 < x < 1} 0 = 
0
$$

but if you arrive from the right, the limit value is $1$,

$$ 
\lim\_{x \to 1 \atop x > 1} \left[ x \right] = 
\lim\_{x \to 1 \atop 2 > x > 1} \left[ x \right] = 
\lim\_{x \to 1 \atop 2 > x > 1} 1 = 
1
$$

So the limit is not defined at $x = 1$, but $f$ is defined there. It is not continous there.