What happens when the basis vectors of an integer lattice are not linearly independent?

The definition of a lattice is that it is a discrete additive subgroup of $\mathbb{R}^n$. The requirement that it be discrete gives us the answer to your question!

The "discrete" bit means that there is some $\epsilon > 0$ such that for any two distinct lattice points $x, y \in \Lambda$, $||x-y|| > \epsilon$.

In English: there has to be some minimum distance by at least which all lattice points are separated.

What would happen if we tried to define a lattice with the following basis:

$\begin{pmatrix} 1 & 0 & \sqrt{2}\\ 0 & 1 & \sqrt{2} \end{pmatrix}$

There's no way we're ever going to get $C_3$ from integer multiples of $C_1$ and $C_2$, so why isn't this a valid lattice basis?

Note that $\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1\end{pmatrix}$ will be in $\Lambda$, as will $\begin{pmatrix} 0.4142135...\\ 0.4142135...\end{pmatrix} = \begin{pmatrix} \sqrt{2} \\ \sqrt{2} \end{pmatrix} - \begin{pmatrix} 1 \\ 1\end{pmatrix}$

as will $ \begin{pmatrix}0.5857864... \\ 0.5857864... \end{pmatrix}= \begin{pmatrix} 1 \\ 1\end{pmatrix} - \begin{pmatrix} 0.4142135...\\ 0.4142135...\end{pmatrix}$ and so on and so forth.

So if I came to you claiming that our set of 3 vectors forms a basis for a 2-D lattice, I would have to offer up some minimum distance between lattice points, $\epsilon$, per the definition of a lattice. However, you would always be able to find two lattice points $x, y \in \Lambda$ such that $|| x - y || < \epsilon$ for any value of $\epsilon$ that I proposed, proving that I was a liar and our set of vectors doesn't form a lattice basis after all.