There are already some great answers given in the comments of the question that answer the question as stated. But just for fun, let's do a proper analysis proof. Also, $\frac1{10}$ is totally arbitrary; we're going to show that $|m+n\sqrt2|$ can always get below any arbitrary positive real $r$.

There is, of course, the trivial solution $m=n=0$, but let's suppose for the sake of argument that you want something else.

With this in mind, let's consider the set $S=\{|m+n\sqrt2|\colon m,n\in\mathbb{Z}, (m,n)\neq(0,0)\}$. Every element in this set is bounded from below (by $0$), and so there must be some real number $r$ which serves as a greatest lower bound for $S$. Let's assume that $r>0$. Then there are two cases: either $r\in S$ or $r\not\in S$.

In the first case, there are integers $m,n$ such that $m+n\sqrt2=r$. (Notice that we can drop the absolute value, because if $m+n\sqrt2=-r$ then $-m-n\sqrt2=r$.) Now, I claim that $S=\{kr\colon k\in\mathbb{Z}\}$, in other words, $S$ is just all of the multiples of $r$. The proof of this is as follows:

Clearly every integer multiple of $r$ is an element of $S$. Suppose that there was some $s=a+b\sqrt 2\in S$ such that $\frac sr\not\in\mathbb{Z}$. Then $\frac sr=k+x$, where $k\in \mathbb{Z}$ and $x\in(0,1)$. So $s=a+b\sqrt2=r(n+x)=kr+rx$. But that means that $s-kr=(a-m)+(b-n)\sqrt2=rx$. Therefore, we have $s-kr\in S$ and $0<s-kr<r$. But this contradicts the fact that $r$ is a greatest lower bound, so such an $s$ cannot exist, and $S$ is the set of integer multiples of $r$.

But we know that $1\in S$ and $\sqrt 2\in S$, and so therefore there are integers $k_1,k_2$ suchthat $1=k_1r$ and $\sqrt2=k_2r$, so $\frac{\sqrt2}1=\sqrt2=\frac{k_1}{k_2}$. So $\sqrt2$ is the ratio of two integers, and therefore $\sqrt2$ is a rational number. This is a contradiction, so we can move on to case $2$.

In our second case, we have $r\not\in S$. But because $r$ is a greatest lower bound of $S$, for every real $\epsilon>0$ there exists some $s\in S$ such that $s\in (r,r+\epsilon)$. (Again, notice that we are justified in dropping the absolute value.) So let $\epsilon=r$, and then we have some $s_1=m_1+n_1\sqrt2\in S$ such that $s_1\in (r,2r)$; in other words, $s_1=r+\epsilon_1$, where $0<\epsilon_1<r$.

But now, we do the same trick, this time letting $\epsilon=\epsilon_1$. What that means is that there is some $s_2=m_2+n_2\sqrt2\in S$ such that $s_2\in (r,r+\epsilon_1)$. So we have $s_2=r+\epsilon_2$, where $0<\epsilon_2<\epsilon_1$. But then, $s_1-s_2=(m_1-m_2)+(n_1-n_2)\sqrt2\in S$, and $s_1-s_2=\epsilon_1-\epsilon_2$. But we know that $0<\epsilon_2<\epsilon_1<r$, and so $0<\epsilon_1-\epsilon_2<r$. In other words, we have again found an element of $S$ which is less than $r$, contradicting the "lower-boundedness" of $r$!

So clearly, such a greatest lower bound cannot exist, and therefore, for any real $r$, we can find integers $m,n$, not both zero, such that $|m+n\sqrt2|<r$. As a special case, letting $r=\frac1{10}$ solves the problem given.