So I'm trying to get a grasp on the separation of variables method - which, as I understand, basically says that $u(x,t) = \varphi(x)G(t)$.
This is how I understand the process:
So given the PDE $\frac{\partial{u}}{\partial{t}} = k\frac{\partial^2{u}}{\partial{x^2}}$, this implies that it can be rewritten as $\frac{\partial}{\partial{t}}(\varphi(x)G(t)) = k\frac{\partial^2}{\partial{x^2}}(\varphi(x)G(t))$, which can be factored to $$\varphi(x)\frac{\mathrm{d}G}{\mathrm{d}t} = kG(t)\frac{\mathrm{d}^2\varphi}{\mathrm{d}x^2}$$
Dividing both sides by $\varphi(x)G(t)$ (as well as by $k$, because I've heard that makes it easier), we get each side as a function of a single variable - the left a function of t, and the right a function of x, which is then set equal to a constant $-\lambda$, since they have to equal the same thing:
$$\frac{1}{kG(t)}\frac{\mathrm{d}G}{\mathrm{d}t} = \frac{1}{\varphi(x)}\frac{\mathrm{d}^2\varphi}{\mathrm{d}x^2} = -\lambda$$
So solving the ODE $\frac{1}{G}\frac{\mathrm{d}G}{\mathrm{d}t} = -k\lambda$, we can write it as $\frac{\mathrm{d}G}{G} = -k\lambda\mathrm{d}t$. Integrating both sides gives the equation $\ln(G)=-k\lambda t + c_1$, and solving for G we get $G = c_1e^{-k\lambda t}$.
To solve the other, I used undetermined coefficients and found that solving the characteristic equation gave complex roots $\pm\sqrt{\lambda}\,i$, leading to the homogeneous solution $\varphi = c_2\cos{\sqrt{\lambda}}+c_3\sin{\sqrt{\lambda}}$.
Then $$u(x,t) = \varphi(x)G(t) = (c_1e^{-k\lambda t} )(c_2\cos{\sqrt{\lambda}}+c_3\sin{\sqrt{\lambda}})$$ $$ = Ae^{-k\lambda t}\cos{\sqrt{\lambda}}+Be^{-k\lambda t}\sin{\sqrt{\lambda}}$$
Where $A = c_1c_2$ and $B = c_1c_3$.
Is this solution correct?
Your solution is correct, but the method that you use doesn't gives the GENERAL solution. It only gives PARTICULAR solutions on the predetermined form $\varphi(x)G(t)$ and on the form of any linear combination of those particular solutions : $$u(x,t)=\sum_{\substack{ any\: \lambda }} A_{(\lambda)} e^{-k\lambda t}\cos(\sqrt{\lambda} \,x) +\sum_{\substack{ any\: \lambda }} B_{(\lambda)} e^{-k\lambda t}\sin(\sqrt{\lambda} \,x) $$ where $A$ and $B$ are any functions of $\lambda$.
These kind of solutions are finite or infinite series.
The above discret form can be extended to : $$u(x,t)=\int A(\lambda) e^{-k\lambda t}\cos(\sqrt{\lambda} \,x)\:d\lambda +\int B(\lambda) e^{-k\lambda t}\sin(\sqrt{\lambda} \,x) \:d\lambda$$ with defined integrals on limited or infinite range.
Also, they are solutions in case of negative $\lambda$, similar to above but with hyperbolic functions instead of circular functions. And even more generally complex solutions with complex $\lambda$.