Consider the IBVP: $u_t - c^{2}u_{xx}=F(t)$ with $u(x,0)=0$ and $u_x(0,t)=u_x(L,t)=0$
The eigenfunctions for this problem is $\phi_{n}(x)=\cos(\frac{k\pi x}{L})$
Thus, we substitute $u(x,t)=\sum_{k=0}a_{k}(t)\phi_{n}(x)$ into our PDE, we get $$\sum_{k=0} [a_{k}'(t)+(\frac{k\pi c}{L})^2 a_{k}(t)]\cos(\frac{k\pi x}{L})=F(t)$$
Then what? I mean.. is this problem really solvable? The left hand side is a sum of product of function of t and x, but the right hand side is just a function of t. Doesn't that means the left hand side has no x dependence? It then follow that the coefficient term $[a_{k}'(t)+(\frac{k\pi c}{L})^2 a_{k}(t)]$ is zero. Then $a_k=Ae^{-\lambda^2 t}$ where $\lambda=\frac{k\pi c}{L}$. But the initial condition $u(x,0)=0$ implies $a_{k}(0)=0$ which then implies $a_{k}(t)=0$ and hence $u(x,t)=0$.
What's wrong with my solution? I don't think it's correct.
You're missing a constant term in your cosine expansion. Remember that $\phi_0(x) = 1$ is also an eigenfunction. Hence
$$ u(x,t) = a_0(t) + \sum_{n=1}^\infty a_n(t)\cos(\lambda_n x) $$
where $\lambda_n = \frac{n\pi}{L}$ are the eigenvalues. Plugging this in, we find
$$ {a_0}'(t) + \sum_{n=1}^\infty \big[{a_n}'(t) - c^2{\lambda_n}^2a_n(t)\big]\cos(\lambda_n x) = f(t) $$
Usually, this is where you'd decompose the inhomogeneous function into its own Fourier series, i.e.
$$ f(x,t) = f_0(t) + \sum_{n=1}^\infty f_n(t)\cos(\lambda_n x) $$
but this is unnecessary since $f(t)$ is constant in $x$, and the cosine series of a constant is that same constant. Thus we can compare coefficients:
\begin{align} {a_0}'(t) &= f(t) \\ {a_n}'(t) - c^2{\lambda_n}^2 a_n(t) &= 0 \end{align}
and initial conditions are all $a_n(0) = 0$. This implies that $a_n(t) = 0$ for $n\ne 0$, and
$$ a_0(t) = \int_0^t f(\tau) d\tau $$
Note that you could also have guessed this solution by finding a particular solution first, then observing that the homogeneous solution has to be $0$.