First of all, note that the limit
$$\int_0^t H(s) \, dB_s = \lim_{n \to \infty} \sum_{i=1}^n H_{t_i} (B_{t_i}-B_{t_{i-1}}) \tag{1}$$
is a limit in $L^2$ (or, alternatively, in probability); the limit does, in general, not exist in a pointwise sense. This means, in particular, that the existence of this limit does not imply that the function $t \mapsto H_t(\omega)$ is Riemann-integrable for each $\omega \in \Omega$.
This is really something you should be careful about when working with such limits: in which sense does the limit exist?
Let's turn to the question why we cannot introduce the Itô integral in a pointwise sense. There is the following general statement which is a direct consequence of the Banach-Steinhaus theorem (see below for a detailed proof):
Theorem Let $\alpha:[a,b] \to \mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := \int_a^b f(t) \, d\alpha(t)$$ exists for all continuous functions $f:[a,b] \to \mathbb{R}$, then $\alpha$ is of bounded variation.
This means that if we define a stochastic integral with respect to a stochastic process $(X_t)_{t \geq 0}$ in a pointwise sense
$$\int_0^t f(s) \, dX_s(\omega), \qquad \omega \in \Omega$$
as a Riemann-Stieltjes integral, then this integral is only well-defined for all continuous functions if $t \mapsto X(t,\omega)$ is of bounded variation (on compacts) for all $\omega \in \Omega$. However, it is well-known that the sample paths of a Brownian motion are almost surely of unbounded variation, and therefore the definition of a stochastic integral in a pointwise sense is not a good idea: the class of functions which we can integrate would not even include the continuous functions.
Proof of the above theorem: The idea for this proof is taken from R. Schilling & L. Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Corollary A.41.
Consider the normed spaces $$(X,\|\cdot\|_X) := (C[a,b],\|\cdot\|_{\infty}) \qquad \text{and} \qquad (Y,\|\cdot\|_Y) := (\mathbb{R},|\cdot|).$$ For a partition $\Pi = \{a=t_0<\ldots<t_n=b\}$ of the interval $[a,b]$ define $I^{\Pi}:X \to Y$ by
$$I^{\Pi}(f) := \sum_{t_j \in \Pi} f(t_{j-1}) (\alpha(t_j)-\alpha(t_{j-1})).$$
Since, by assumption, $I^{\Pi}(f) \to I(f)$ as $|\Pi| \to 0$ for all $f \in C[a,b]$, we have
$$\sup_{\Pi} |I^{\Pi}(f)| \leq c_f < \infty$$
for all $f \in C[a,b]$. Applying the Banach Steinhaus theorem we find
$$\sup_{f \in C[a,b], \|f\|_{\infty} \leq 1} \sup_{\Pi} I^{\Pi}(f) < \infty$$
which is equivalent to
$$\sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty. \tag{2}$$
For a partition $\Pi$ let $f_{\Pi}:[a,b] \to \mathbb{R}$ be a piecewise linear continuous function such that
$$f_{\Pi}(t_{j-1}) = \text{sgn} \, (\alpha(t_j)-\alpha(t_{j-1}))$$
for all $j=1,\ldots,n$. By the very choice of $f_{\Pi}$, we have
$$I^{\Pi}(f_{\Pi}) = \sum_{t_j \in \Pi} |\alpha(t_j)-\alpha(t_{j-1}))| \leq \sup_{f \in X, \|f\|_X \leq 1} |I^{\Pi}(f)|. \tag{3}$$
Combining $(2)$ and $(3)$, we conclude
$$\text{VAR}_1(\alpha,[a,b]) = \sup_{\Pi} I^{\Pi}(f_{\pi}) \leq \sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty.$$
Such iterated integrals are closely related to Hermite polynomials.
If we define
$$H_n(t,x) = \frac{(-t)^n}{n!} \exp \left( \frac{x^2}{2t} \right) \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right),$$
then a straight-forward (but messy) computation shows that
$$dH_{n+1}(t,W_t) = H_n(t,W_t) \, dW_t \tag{1}$$
see this answer for details. You can easily compute the first Hermite polynomials $H_n$: $$H_1(t,x)=x \qquad H_2(t,x) = \frac{x^2}{2}- \frac{t}{2} \qquad \ldots.$$
Claim: $$I(n) = H_n(t,W_t). \tag{2}$$
We prove the assertion by induction. For $n=1$ it is obvious that $I(1)=W_t = H_1(t,W_t)$. Now suppose that $(2)$ holds for $n=k$, then we obtain from $(1)$ that $$H_{k+1}(t,W_t) = \int_0^t H_k(s,W_s) \, dW_s \stackrel{\text{ind. hypothesis}}{=} \int_0^t \left( \int_0^s \dots \int_0^{s_2} dW_{s_1} \ldots \, dW_{s_k} \right) \, W_s,$$ i.e. $(2)$ holds for $n=k+1$.
Best Answer
The trick is to rewrite the sum to
$$\sum_{j=1}^{k_{n}} W_{t_{j-1}} \left(W_{t_{j}}-W_{t_{j-1}}\right) +\sum_{j=1}^{k_{n}} (W_{t^{\lambda}_{n,j}}-W_{t_{j-1}})^{2} + \sum_{j=1}^{k_{n}} (W_{t^{\lambda}_{n,j}}-W_{t_{j-1}}) \left(W_{t_{j}}-W_{t^{\lambda}_{n,j}}\right).$$
Then the first sum is clear, the second is almost a quadratic variation (gives $\lambda T$) and the last one goes to 0.