What is meant by a continuous-time white noise process?

This is a bit late, but I see that the main points in this question have not been completely addressed. I'll set \begin{equation} \sigma = 1 \end{equation} for this answer.

The definition of white noise may be context-dependent: How you define it depends on what you want to do with it. There's nothing inherently wrong with saying that white noise (indexed by a set $T$) is just the process of iid standard normal random variables indexed by $T$, i.e. $E[X(t)X(s)] = \begin{cases} 1 & t = s \\ 0 & t \neq s \end{cases}.$ However, as cardinal noted here, Example 1.2.5 of Kallianpur's text shows that this process is not measurable (as a function of $(t, \omega)$). This is why, as Did commented above, $Y$ is undefined (with this definition of $X$). Thus, this definition of white noise is not appropriate for defining objects like $Y$.

Rather, you want $X$ to have covariance given by the Dirac delta. But the $\delta$ function is not a function but rather a measure and the best context for understanding it is the theory of distributions (or generalized functions---these are not to be confused with "probability distributions"). Likewise, the appropriate context for white noise is the theory of random distributions.

Let's warm up with a heuristic explanation: We'll think of white noise as the "derivative" of Brownian motion: "$dB_t/dt = X_t$". So ignoring rigor for a moment, we could write \begin{equation} \int_0^T h(t) X(t) dt = \int_0^T h(t) \frac{dB_t}{dt} dt = \int_0^T h(t) dB_t. \end{equation}

The reason this isn't rigorous is that Brownian motion is nowhere differentiable. However, the theory of distributions allows us to "differentiate" non-differentiable functions. First of all, a distribution is a linear functional (linear map taking values in the real numbers) on a space of "test functions" (usually smooth functions of compact support). A continuous function $F$ can be viewed as a distribution via the pairing \begin{equation} (F, f) = \int_0^\infty F(t) f(t) dt. \end{equation} The distributional derivative of $F$ is the distribution $F'$ whose pairing with a test function $f$ is defined by \begin{equation} (F', f) = -(F, f'). \end{equation}

Thinking of Brownian motion as a random function, we can define white noise $X$ as its distributional derivative. Thus, $X$ is a random distribution whose pairing with a test function $f$ is the random variable \begin{equation} (X, f) = -(B, f') = -\int_0^\infty B(t) f'(t) dt. \end{equation} By stochastic integration by parts, \begin{equation} (X, f) = \int_0^\infty f(t) dB_t; \end{equation} this is the Itô integral of $f$ with respect to $B$.

Now a well-known fact in stochastic calculus is that $M_T = \int_0^T f(t) dB_t$ is a martingale starting at $M_0 = 0$, so $E (X, f) = 0$. Moreover, by the Itô isometry, \begin{equation} \mathrm{Var}((X, f)) = E (X, f)^2 = \int_0^\infty f(t)^2 dt. \end{equation} It can also be verified that $(X, f)$ is Gaussian.

My main point is that a more appropriate definition of $Y$ might be \begin{equation} Y = \int_0^T h(t) dB_t. \end{equation}

As a last note, because of the way $X$ was defined above, $X_t$ is not defined but $(X, f)$ is. That is, $X$ is a stochastic process but whose index set is given by $T = \{ \text{test functions} \}$ rather than $T = [0, \infty)$. Moreover, again by the Itô isometry, \begin{equation} E (X, f) (X, g) = \int_0^\infty f(t) g(t) dt. \end{equation} Abandoning rigor again, this becomes \begin{equation} E (X, f) (X, g) = \int_0^\infty \int_0^\infty f(s) \delta(s - t) g(t) ds dt \end{equation} and it is in this sense that the covariance of $X$ is the Dirac delta.

Edit: Note that we could leave the definition of $(X, f)$ in terms of the ordinary integral and do all the above calculations using Fubini's theorem and (ordinary) integration by parts (it's just a bit messier).