Say $a_n>0$ and $\sum a_n<\infty$. Define $$u(z)=\sum a_n\log|z-1/n|.$$
Then $u$ is subharmonic; you could verify this directly or note that it's the logarithmic potential of a finite measure. But if $a_n\to0$ fast enough then $u(0)>-\infty$, hence $u$ is not continuous.
(If I have my inequalities straight then you can get a real-valued example by considering $\phi\circ u$ for an appropriate convex function $\phi$, for example $\phi(t)=e^t$.)
Edit: To answer questions that came up in the comments: No, there's no such example on $\Bbb R$, because there subharmonic is the same as convex, and convex functions are continuous.
Why is convexity equivalent to subharmonicity on $\Bbb R$? We don't need to cite Seirpinski, it's easy. Of course $u$ is convex if $$u(tx+(1-t)y)\le tu(x)+(1-t)u(y)\quad(0\le t\le 1)\quad(*).$$
The defininig inequality for subharmonic functions is $(*)$ for $t=1/2$, "midpoint convex". Since for example $$\frac12(x+\frac12(x+y))=\frac34x+\frac14y$$it's easy to see that midpoint convexity implies (*) for dyadic rationals $t$. But we're not assuming $u$ is continuous...
Heh: Recall that if $u$ is subharmmonic, $v$ is continuous in a closed disk and harmonic in the interior, and $u\le v$ on the boundary then $u\le v$ in the disk. The proof of that works just as well in one variable, and if you think about it you see that it says exactly that a subharmonic function on $\Bbb R$ satisfies $(*)$.
So, on the line, midpoint convex plus usc, ie subharmonic, implies convex. The converse is clear.