For Red, it is not true that neither of his pure strategies dominates the other. It is true that neither of his pure strategies strictly dominates the other, but since
$$\ R\left(\mbox{Up, Left}\right) =-1 \ge R\left(\mbox{Up, Right}\right)=-1\ ,\ \mbox{and}\\
R\left(\mbox{Down, Left}\right)=2 > R\left(\mbox{Down, Right}\right)=-1\ ,$$
where $\ R\ $ is Red's payoff matrix, then his first pure strategy, Left, clearly dominates his second, Right. And since
$$ B\left(\mbox{Up, Left}\right) = 3 > B\left(\mbox{Down, Left}\right)=-1\ ,$$
where $\ B\ $ is Blue's payoff matrix, then $\ \mbox{(Up, Left)}\ $ is a pure strategy Nash equilibrum.
Response to comment: Mixed strategy Nash equilibria for general bimatrix games can be found with the Lemke-Howson algorithm.
However, for the simple game described here (with Red's payoff matrix modified as proposed in the comment), it's possible to find a mixed strategy Nash equilibrium simply by solving a couple of linear equations. Blue's payoff matrix is
$$ B=\begin{pmatrix}
3&-1\\
-1&2
\end{pmatrix}\ ,
$$
so if Red plays the mixed strategy $\ \left(q, 1-q\right)\ $, Blue's expected payoff will be
$$ 3q-(1-q)$$
when he chooses his first pure strategy (Up), and
$$-q+2(1-q)$$
when he chooses his second pure strategy (Down). Thus, if Red chooses $\ q\ $ to make both these expected payoffs equal,
$$ 3q-(1-q)=-q+2(1-q)\ ,$$
$\ \left(\ q=\frac{3}{7}\ \right)\ $, then Blue's expected payoff will be $\ 3\cdot \frac{3}{7}- \frac{4}{7}=\frac{5}{7}=-\frac{3}{7}+2\cdot\frac{4}{7}\ $ regardless of what strategy (whether pure or mixed) he chooses.
Likewise, Red's payoff matrix (with the amendment proposed in the comment) is
$$
R=\begin{pmatrix}
-2&-1\\
2&-1
\end{pmatrix}\ ,
$$
so if Blue chooses a mixed strategy $\ \left(p, 1-p\right)\ $ such that
$$ -2p+2\left(1-p\right)=-p-\left(1-p\right)\ ,$$
that is, with $\ p=\frac{3}{4}\ $, then Red's expected payoff will be $-2\cdot\frac{3}{4}+2\cdot\frac{1}{4}\ = -1 = -\frac{3}{4}-\frac{1}{4}\ $ regardless of what strategy (whether pure or mixed) he chooses.
Thus, the mixed strategies $\ \left(\frac{3}{4}, \frac{1}{4}\right)\ $ for Blue, and $\ \left(\frac{3}{7}, \frac{4}{7}\right)\ $ for Red constitute a Nash equilibrium for this gane.
