Irreducible representations of $C(T,B(X))$ Let $T$ be a compact topological space, $X$ a finite-dimensional Hilbert space, $B(X)$ the algebra of operators in $X$, and $C(T,B(X))$ the $C^*$-algebra of continuous maps from $T$ into $B(X)$ (with the poinwise algebraic operations and the uniform norm). I think, all irreducible representations of $C(T,B(X))$ must be of the form
$$
\pi:C(T,B(X))\to B(X),\qquad \pi(f)x=f(t)x,\qquad x\in X,\ f\in C(T,B(X)),
$$
for some $t\in T$.
How is this proved?

Strictly speaking, the answer is no. The following is true: every irreducible representation is unitarily equivalent to a "point-evaluation". That is, there exists $t\in T$ and a unitary operator $U\in B(X)$ such that 
$$
\pi:C(T,B(X))\to B(X),\qquad \pi(f)x=(U^*f(t)U)x,\qquad x\in X,\ f\in C(T,B(X)).
$$
It follows from a general theorem for continuous fields of $C^*$-algebras, see e.g. the book by J.Dixmier "$C^*$-algebras and representation theory", Corollary 10.4.4.
You can prove it yourself by showing (similarly to the case $C(T,\mathbb C)\ $) that every closed two-sided ideal $I\subseteq C(T,B(X))$ is of the form 
$$
I_S=\{f\in C(T,B(X))\ \ \colon\  f|_S\equiv 0\}
$$ 
for some closed $S\subseteq T.$ The kernel of an irreducible representation is then $I_{\{t\}}$ for some $t\in T.$ 
