NB: in the following, it is assumed that $X, Y, Z$ and $R$ are continuous.
If $C \subset TM$ is the compact subset from which $X$, $Y$ and $Z$ are taken, then $C \times C \times C$ is compact (Tychonoff's theorem), and the map
$R(X, Y)Z: C \times C \times C \to TM \tag{1}$
is continuous; this follows by virtue of the fact that in any local coordinate system $w_\alpha$ on $M$, with corresponding tangent basis vector fields $e_\alpha$ where
$e_\alpha = \dfrac{\partial}{\partial w_\alpha}, \tag{2}$
we may write
$R(X, Y)Z = e_\beta R^\beta_{\mu \nu \rho} X^\mu Y^\nu Z^\rho, \tag{3}$
where
$X = X^\mu e_\mu \tag{4}$
and so forth. (The Einstein summation convention is understood as being operative in (3) and (4), as throughout this answer.) Since all of the coefficients $R^\beta_{\mu \nu \rho}$, $X^\mu$ and so forth are continuous, as well as the vector fields $e_\alpha$, the continuity of $R(X, Y)Z$ follows. Note that we are in essence drawing on the fact that a composition of continuous maps is itself continuous.
Denoting the metric norm by $\Vert \cdot \Vert_g$, we see in a manner similar to the above that
$\Vert \cdot \Vert_g: TM \to \Bbb R, \tag{5}$
which takes tangent vectors $X \in TM$ to their metric norms $\Vert X \Vert_g = (g((X, X))^{1/2}$, is also continuous. It then further follows that the composed map
$\Vert R(X, Y)Z \Vert_g: C \times C \times C \to TM \to \Bbb R \tag{6}$
is continuous, again since it is a composition of continuous functions.
It now follows from the compactness of $C \times C \times C$ that $\Vert R(X, Y)Z \Vert_g$ is bounded on this set, since a continuous function on a compact set is always bounded.
Note Added Friday 26 June 2015 1:47 PM PST: In the light of our colleague Thomas well-framed answer and comments, as well as those of our OP Mathmath, I don't suppose it's necessary to say much more, so I will forgo adding another answer to address the case in which $X$, $Y$, $Z$ need not be continuous; but I would like to take this opportunity add a few (hopefully brief) words on the subject. I was indeed aware that a bound should exist whether or not the continuity of $X$, $Y$, $Z$ binds, that that it would most likely take the form
$\Vert R(X, Y)Z \Vert_g \le \Vert R \Vert \Vert X \Vert_g \Vert Y \Vert_g \Vert Z \Vert_g, \tag{7}$
where $\Vert R \Vert$ may be taken to be some constant depending only on the tensor field $R$. In fact, (7) holds pointwise, without any reference to continuity at all. Whilst thinking about how to best write-up an explanation of (7), it occurred to me that in the case of $X$, $Y$, $Z$ continuous one can altogether dispense with the details of multilinearity and argue, more or less directly, from the fact that $R(X, Y)Z$ is a polynomial in the components of $X$, $Y$, $Z$ with continuous coefficients the $R^\alpha_{\mu \nu \rho}$; thus, $\Vert R(X, Y)Z \Vert_g$ is a continuous mapping on the compact set $C \times C \times C$. And having spied that ball, I picked it up and ran with it, forgetting in my haste that continuity of $X$, $Y$, $Z$ is not essential here. I must confess that I am not used to considering vector fields sans continuity, so I was pre-disposed to read it into the hypotheses of this question; but still, I think it worth noting that the general continuity argument can, in certain situations, replace the more linearity-specific methods which depend on estimates such as (7). End of Note.
