This question is a follow-up from the following: conditional expectation of stochastic integral so I won't repeat myself regarding assumptions and notation.
Using Brownian bridge approach, we know that ${\mathbb E}[W_t|W_T]=\frac{t}{T}W_T$. This is compatible with a regression decomposition of $W_t$ on $W_T$, such as:
$$ W_t = \beta^W_t W_T + \epsilon $$ for t $\leq T$, where $\epsilon \sim \mathcal{N}(0,1)$ is an independent noise and $\beta^W_t$ can be interpreted as a standard OLS estimator, indeed
$$ \beta^W_t = \frac{{\mathbb Cov}(W_t,W_T)}{{\mathbb Var}(W_T)} = \frac{{\mathbb E}[W_t W_T]}{{\mathbb E}[W^2_T]} = \frac{t}{T} $$
In question conditional expectation of stochastic integral, we showed that the conditional expectation of the stochastic integral of a deterministic function $\sigma_t$ $$ M_t = \int_0^t \sigma_s dW_s $$ w.r.t. to the Wiener process at $T \geq t$ can be written as
$$ {\mathbb E}[M_t|W_T] = \frac{\int^t_0 \sigma_s ds}{T} W_T $$
By analogy, we extend the above regression decomposition as
$$ M_t = \beta^M_t W_T + \epsilon $$
with
$$ \beta^M_t = \frac{\int^t_0 \sigma_s ds}{T} $$
Now, $\beta^M_t$ can be properly interpreted as an OLS estimator as long as
$$ \beta^M_t = \frac{{\mathbb Cov}(M_t,W_T)}{{\mathbb Var}(W_T)} = \frac{\int^t_0 \sigma_s ds}{T} $$
that is to say, as long as the covariance between the stochastic integral $M_t$ and the Wiener $W_T$ is
$$ {\mathbb Cov}(M_t,W_T) = \int^t_0 \sigma_s ds $$
which is the conjecture we'd like to prove.