First, since the set of $w_n$ is orthonormal, you get (note that the scalar product is continuous!) $$
\langle x,w_k \rangle
= \underbrace{\langle v,w_k \rangle}_{=0} + \sum_{i=1}^\infty a_n \underbrace{\langle w_i,w_k\rangle}_{=\delta_{i,k}}
= a_n \text{.}
$$
From Hilbert-Schmidt and the boundedness (i.e. continuity!) of $T$ it follows that $$
T(x)
= T\left(v + \sum_{k=1}^\infty a_nw_n\right)
= \underbrace{T(v)}_{=0} + \sum_{k=1}^\infty
\underbrace{a_n}_{=\langle x,w_n\rangle}
\underbrace{T(w_n)}_{=\lambda_n w_n}
= \sum_{k=1}^\infty \lambda_n\langle x,w_n\rangle w_n
\text{.}
$$
All that remains is to find an orthonormal basis $B$ of $H$ with $B \supset \{w_n\}$. You get such a thing by picking some orthonormal basis $\{v_n\}$ of $\mathcal{N}(T)$, and setting $B = \{w_n\} \cup \{v_n\}$. This works because $H$ is the direkt sum of $\mathcal{N}(T)$ and $\mathcal{N}(T)^\bot$, since $\mathcal{N}(T)$ is closed.
OK. I got the uniformity I needed. Here's how. In the process of proving that $E$
is a resolution of the identity, I also proved that the sum
$$\sum_\alpha\lvert\nu_\alpha\rvert(e)\qquad(e\in\mathscr{B}_T)$$
converges.
Split $f_1$ into four positive parts:
real positive, real negative, imaginary positive and imaginary negative. Each part
is bounded on $\sigma(T)$ so there is a sequence of simple measurable functions
converging to the part uniformly on $\sigma(T)$. The sum of these four sequences
is then a sequence $\{s_n\}$ of simple measurable functions converging to $f_1$
uniformly on $\sigma(T)$.
If $\epsilon>0$ is given, then there exists $N$ such that for all $n\geq N$,
and for all $\lambda\in\sigma(T)$,
\begin{equation*}
\lvert f_1(\lambda)-s_n(\lambda)\rvert
<\frac{\epsilon}{\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))}.
\end{equation*}
Then
\begin{equation*}
\begin{split}
\Biggl\lvert\int_{\sigma(T)}\!(f_1-s_n)\,
d({\textstyle \sum_{i=1}^k}\nu_{\alpha_i})\Biggr|
&\leq\int_{\sigma(T)}\!\lvert f_1-s_n\rvert\,
d(\lvert{\textstyle\sum_{i=1}^k}\nu_{\alpha_i}\rvert)\\
&<\epsilon\frac{\lvert\sum_{i=1}^k\nu_{\alpha_i}\rvert(\sigma(T))}
{\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))}
\leq\epsilon\frac{\sum_{i=1}^k\lvert\nu_{\alpha_i}\rvert(\sigma(T))}
{\sum_\alpha\lvert\nu_\alpha\rvert(\sigma(T))}
\leq\epsilon,
\end{split}
\end{equation*}
for all $n\geq N$ and for all $k=1,2,\dots$.
Therefore,
\begin{equation}\tag{3}\label{3}
\lim_{n\to\infty}\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i})
=\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i})
\end{equation}
uniformly for $k=1,2,\dots$. \eqref{3} also holds for $k=\infty$.
Now using \eqref{3}, the left hand side of (1) can be rewritten as
\begin{equation*}
\sum_\alpha\int_{\sigma(T)}\!f_1\,d\nu_\alpha
=\lim_{k\to\infty}\sum_{i=1}^k\int_{\sigma(T)}\!f_1\,d\nu_{\alpha_i}
=\lim_{k\to\infty}\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i})
=\lim_{k\to\infty}\lim_{n\to\infty}\int_{\sigma(T)}\!s_n\,
d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i}),
\end{equation*}
and we can interchange the limits, since the $n$ limit is uniform for $k$ and the
iterated limit exists. Therefore, interchanging the limits we get
\begin{equation*}
\begin{split}
\sum_\alpha\int_{\sigma(T)}\!f_1\,d\nu_\alpha
&=\lim_{n\to\infty}\lim_{k\to\infty}\int_{\sigma(T)}\!s_n\,
d({\textstyle\sum_{i=1}^k}\nu_{\alpha_i})
=\lim_{n\to\infty}\lim_{k\to\infty}\sum_{i=1}^k
\int_{\sigma(T)}\!s_n\,d\nu_{\alpha_i}\\
&=\lim_{n\to\infty}\sum_\alpha\int_{\sigma(T)}\!s_n\,d\nu_\alpha
=\lim_{n\to\infty}\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_\alpha}\nu_\alpha)
=\int_{\sigma(T)}\!f_1\,d({\textstyle\sum_\alpha}\nu_\alpha),
\end{split}
\end{equation*}
which is (1). The last equality is because \eqref{3} holds for $k=\infty$.
The justification for the next to last equality can be found in
Halmos Introduction to Hilbert Space and the Theory of Spectral Multiplicity as
Theorems 7.1 and 7.2, remembering that if
$s_n=a_1\chi_{e_1}+\cdots+a_j\chi_{e_j}$, then
\begin{equation*}
\begin{split}
\sum_\alpha\int_{\sigma(T)}\!s_n\,d\nu_\alpha
&=\sum_\alpha(a_1\nu_\alpha(e_1)+\cdots+a_j\nu_\alpha(e_j))
=a_1\sum_\alpha\nu_\alpha(e_1)+\cdots+a_j\sum_\alpha\nu_\alpha(e_j)\\
&=a_1({\textstyle\sum_\alpha}\nu_\alpha)(e_1)
+\cdots+a_j({\textstyle\sum_\alpha}\nu_\alpha)(e_j)
=\int_{\sigma(T)}\!s_n\,d({\textstyle\sum_\alpha}\nu_\alpha).
\end{split}
\end{equation*}
Best Answer
More generally, any $^{\ast}$-homomorphism $f : A \to B$ between (edit: unital) C*-algebras is automatically continuous.
Proof. This is a standard argument. If $a - \lambda I$ is invertible then so is $f(a - \lambda I) = f(a) - \lambda I$; it follows that the spectrum satisfies $\sigma(f(a)) \subseteq \sigma(a)$, so the spectral radius satisfies $\rho(f(a)) \le \rho(a)$. Now, for every $a \in A$ we have
$$\| f(a) \| = \sqrt{ \| f(a)^{\ast} f(a) \| } = \sqrt{ \rho(f(a^{\ast} a)) } \le \sqrt{ \rho(a^{\ast} a)} = \| a \|$$
so $f$ has norm $\le 1$ and in particular is continuous. The use of the C*-identity here is what allows us to use the spectral radius even though $a$ and $f(a)$ are not guaranteed to be normal. $\Box$