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: 
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 $[0,2]$.
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.