Show that $M_2(\mathbb{R})$ has no non-trivial two-sided ideals In addition to the title question, I also want to find a non-trivial right ideal and a non-trivial left ideal of $M_2(\mathbb{R})$ .
Attempt of title question:
Suppose $\exists I\subset M_2(\mathbb{R})$ s.t. $\forall x\in I, \forall r\in M_2(\mathbb{R})$, $rxr\in I$ $\Rightarrow $ $rx\in I, xr\in I$, but $M_2(\mathbb{R})$ is a non-commutative ring. So, w.l.o.g. if $rx\in I$ then $xr\not\in I$. Contradiction. I have an inkling this is mistaken.
I am also confused about the left and right ideals, but an idea that seems to work for a right ideal is to let $x=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$. A left ideal would be similar.
 A: Your example for a left and right ideal is fine. Your argument for the other part is not right: noncommutative rings can have nontrivial ideals. (I can't think of a very simple example off the top of my head, but it's true.)
To show that $M_2(\mathbb R)$ has no nontrivial ideals, assume a nonzero element $A=\big(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big)$ is contained in a nontrivial ideal. You can multiply $A$ by elementary matrices (i.e. perform row and column operations) to reduce it to one of the three matrices
$$
\begin{bmatrix}0&0\\0&0\end{bmatrix},\,
\begin{bmatrix}1&0\\0&0\end{bmatrix},\,
\begin{bmatrix}1&0\\0&1\end{bmatrix}.
$$
Now what can we conclude? (I omitted a lot of details. Can you fill them in?)
A: Assuming $M_2(\mathbb{R})$ refers to the set of all $2 \times 2$ real matrices:
If $A$ is a rank $2$ matrix, then $(A) = (I)$ is the whole ring.  If $A$ is the zero matrix, then we get the trivial ideal.  The only remaining possibility is that $A$ is a rank $1$ matrix.
By performing row operations (i.e. multiplying on the left by units) and column operations (i.e. multiplying on the right by units), we can produce any other rank $1$ matrix.  So, we have
$\pmatrix{1&0\\0&0} \in (A)$ and $\pmatrix{0&0\\0&1} \in (A)$.  Because $(A)$ is closed under addition, $I \in (A)$, which means that $(A)$ is the entire ring.
Thus, the only two two-sided ideals are the trivial ideal and the ring itself.
As for nontrivial one-sided ideals, consider
$$
\left\{A\pmatrix{1&0\\0&0}: A \in M_2(\mathbb{R})\right\}\\
\left\{\pmatrix{1&0\\0&0}A: A \in M_2(\mathbb{R})\right\}
$$ 
A: For your question on left/right ideals:
By "let $x= \begin{pmatrix}1 &1\\0 &0\end{pmatrix}$" do you mean the right ideal generated by $x$? (Obviously an idea is not equal to an element)? Of course this is true, given any element of a ring, the right ideal it generates is by definition a right ideal.
In more generality, simple submodules of $M_2(\mathbb R)$ (as a right module over itself) are isomorphic to $\mathbb R^2$ - corresponding to nonzero rows. Thus we have two simple submodules:
$$ M_1 = \left\{\begin{pmatrix}\alpha &\beta \\0 &0\end{pmatrix} : \alpha , \beta \in \mathbb R \right\}$$
$$M_2 = \left\{\begin{pmatrix}0 &0\\\gamma &\delta\end{pmatrix} : \gamma , \delta \in \mathbb R \right\}$$
Can you show this? And moreover, now categorise all right ideals?
Similar arguments will show that all left ideals correspond to (direct sums of) columns in $\mathbb R^2$
