The solution of the heat equation $u_t=\alpha u_{xx}$ is continuous on time $[0,\infty)$ Consider the heat equation with Dirichlet boundary condition on a metal rod of length $l$
\begin{equation}
\frac{\partial}{\partial t}u(t,x)=\alpha\frac{\partial^{2}}{\partial x^{2}}u(t,x)\label{eq:edp}
\end{equation}
\begin{equation}
u(t,0)=0\,\,\,\,\,\text{et}\,\,\,\,\,u(t,l)=0\label{eq:bords}
\end{equation}
\begin{equation}
u(0,x)=f(x)\label{eq:condition initiale}
\end{equation}
Using separation of variable method we can solve this system and we get $$u(t,x)=\sum_{n=1}^{\infty}b_{n}e^{-\alpha\left(\frac{\pi n}{l}\right)^{2}t}\sin\left(\frac{\pi n}{l}x\right),$$where $$b_n=\frac{2}{l}\int_0^lf(y)\sin\left(\frac{\pi n}{l}y\right)dy.$$
Using the M-test of uniform convergence, we can prove that for each $x\in[0,l]$ the time function $t\mapsto u(t,x)$ is continuous on all intervals $[\epsilon,\infty)$ thus continuous on $(0,\infty)$. Moreover, we can prove that for each $t>0$ the space function $x\mapsto u(t,x)$ is continuous on $[0,l]$ no matter how the function $f$ is ($f(x)$ is hidden inside the $b_n$ in the series formula and does not have a role in the $x$  variable )
I am just curious about the continuity of the time function $t\mapsto u(t,x)$ on $[0,\infty)$. For an initial date such as $f(x)=\sin\left(\frac{\pi }{l}x\right)$, we have this continuity since in this case we avoid the mystery of infinities and we have simply $$u(t,x)=be^{-\alpha\left(\frac{\pi}{l}\right)^{2}t}\sin\left(\frac{\pi }{l}x\right).$$
What can we say for some general initial data $f(x)$?
 A: You have shown that the series converges uniformly for $(t,x) \in [\delta,\infty)\times[0,l]$ where $\delta > 0$ and, thus,  $t \mapsto u(t,x)$ is continuous for all $t > 0$.  If we can extend the uniform convergence for $t \in [0,\infty)$, then continuity at $t=0$ would follow, i.e., $\lim_{t \to 0+}u(t,x) = u(0,x) = f(x)$ for all $x \in [0,l]$.
Given  that the Fourier series $$f(x) = u(0,x)= \sum_{n=1}^\infty b_n \sin \frac{\pi n x}{l}$$ is  convergent, it is uniformly convergent, trivially, for $t \in [0,\infty)$. We also have that $e^{-\alpha n^2\pi^2t/l^2}$ is monotone decreasing  with respect to $n$, and, uniformly bounded for all $t\in [0,\infty)$ since $|e^{-\alpha n^2\pi^2t/l^2}| \leqslant 1$.  By Abel's test for uniform convergence of series, it follows that (with $x$ fixed) the series
$$u(t,x)= \sum_{n=1}^\infty b_n e^{-\alpha n^2\pi^2t/l^2}\sin \frac{\pi n x}{l}$$
is uniformly convergent for all $t \in [0,\infty)$, and, therefore, $t \mapsto u(t,x)$ is continuous at $t = 0+$.
