unitary operators in Hilbert's space are arcwise connected From books, Murphy: $C^\star$-algebras and operator theory, and Douglas: Banach Algebra Techniques in Operator Theory.
Both show that unitary operators in Hilbert space are arcwise connected, using functional calculus or borel functional calculus.
And my question... is posible show it, without use functional calculus, a more elementary proof. any book do it for read.
thanks
 A: I'm not aware of any easier proof, and I doubt there even is an elementary proof for the following reason:
If one considers the finite-dimensional case, the standard argument (as you may know) is that any unitary $U\in\mathbb C^{n\times n}$ due to the spectral theorem can be written as $U=V\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})V^*$ for some $V$ unitary and $\theta_1,\ldots,\theta_n\in\mathbb R$. Then the map
$$
f:[0,1]\to\mathbb C^{n\times n}\qquad t\mapsto V\operatorname{diag}(e^{it\theta_1},\ldots,e^{it\theta_n})V^*=e^{it(V\operatorname{diag}(\theta_1,\ldots,\theta_n)V^*)}\tag{1}
$$
is continuous with $f(t)$ unitary for all $t\in[0,1]$, $f(1)=U$ and $f(0)=\operatorname{id}$ which shows connectedness of the unitary group. The idea of this proof is not all-too surprising, given the unitary group is a Lie group so it is intrinsically connected to the exponential map.
Now to lift this proof to infinite dimensions you need to be able to say something about the eigenvalues of a unitary operator $U\in\mathcal B(\mathcal H)$ and/or write it as $e^{iA}$ for some $A\in\mathcal B(\mathcal H)$ self-adjoint as done in (1). But for this you need (Borel) functional calculus, and I don't see how one can find a way around using the exponential map to somewhat explicitely construct a path---even in finite dimensions where everything is easier by default. 
